Calculator search results

$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 30 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 56 } = 0$

$acx^{2} + \left(ad + bc\right)x +bd = \left(ax + b\right)\left(cx+d\right)$

$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 28 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) = 0$

$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 28 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) = \color{#FF6800}{ 0 }$

$ $ If the product of the factor is 0, at least one factor should be 0 $ $

$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 28 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } = \color{#FF6800}{ 0 } \end{array}$

$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 28 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } = \color{#FF6800}{ 0 } \end{array}$

$ $ Solve the equation to find $ x$

$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 28 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ 2 } \end{array}$

$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 30 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 56 } = \color{#FF6800}{ 0 }$

$ $ Convert the quadratic expression on the left side to a perfect square format $ $

$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 15 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 56 } \color{#FF6800}{ - } \color{#FF6800}{ 15 } ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$

$\left ( x - 15 \right ) ^ { 2 } \color{#FF6800}{ + } \color{#FF6800}{ 56 } \color{#FF6800}{ - } \color{#FF6800}{ 15 } ^ { \color{#FF6800}{ 2 } } = 0$

$ $ Move the constant to the right side and change the sign $ $

$\left ( x - 15 \right ) ^ { 2 } = \color{#FF6800}{ - } \color{#FF6800}{ 56 } \color{#FF6800}{ + } \color{#FF6800}{ 15 } ^ { \color{#FF6800}{ 2 } }$

$\left ( x - 15 \right ) ^ { 2 } = - 56 + \color{#FF6800}{ 15 } ^ { \color{#FF6800}{ 2 } }$

$ $ Calculate power $ $

$\left ( x - 15 \right ) ^ { 2 } = - 56 + \color{#FF6800}{ 225 }$

$\left ( x - 15 \right ) ^ { 2 } = \color{#FF6800}{ - } \color{#FF6800}{ 56 } \color{#FF6800}{ + } \color{#FF6800}{ 225 }$

$ $ Add $ - 56 $ and $ 225$

$\left ( x - 15 \right ) ^ { 2 } = \color{#FF6800}{ 169 }$

$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 15 } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 169 }$

$ $ Solve quadratic equations using the square root $ $

$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 15 } = \pm \sqrt{ \color{#FF6800}{ 169 } }$

$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 15 } = \pm \sqrt{ \color{#FF6800}{ 169 } }$

$ $ Solve a solution to $ x$

$\color{#FF6800}{ x } = \pm \color{#FF6800}{ 13 } \color{#FF6800}{ + } \color{#FF6800}{ 15 }$

$\color{#FF6800}{ x } = \pm \color{#FF6800}{ 13 } \color{#FF6800}{ + } \color{#FF6800}{ 15 }$

$ $ Separate the answer $ $

$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 15 } \color{#FF6800}{ + } \color{#FF6800}{ 13 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ 15 } \color{#FF6800}{ - } \color{#FF6800}{ 13 } \end{array}$

$\begin{array} {l} x = \color{#FF6800}{ 15 } \color{#FF6800}{ + } \color{#FF6800}{ 13 } \\ x = 15 - 13 \end{array}$

$ $ Add $ 15 $ and $ 13$

$\begin{array} {l} x = \color{#FF6800}{ 28 } \\ x = 15 - 13 \end{array}$

$\begin{array} {l} x = 28 \\ x = \color{#FF6800}{ 15 } \color{#FF6800}{ - } \color{#FF6800}{ 13 } \end{array}$

$ $ Subtract $ 13 $ from $ 15$

$\begin{array} {l} x = 28 \\ x = \color{#FF6800}{ 2 } \end{array}$

$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 30 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 56 } = \color{#FF6800}{ 0 }$

$ $ Solve the quadratic equation $ ax^{2}+bx+c=0 $ using the quadratic formula $ \dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 30 } \right ) \pm \sqrt{ \left ( \color{#FF6800}{ - } \color{#FF6800}{ 30 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \color{#FF6800}{ 56 } } } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } } }$

$x = \dfrac { \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } 30 \right ) \pm \sqrt{ \left ( - 30 \right ) ^ { 2 } - 4 \times 1 \times 56 } } { 2 \times 1 }$

$ $ Simplify Minus $ $

$x = \dfrac { 30 \pm \sqrt{ \left ( - 30 \right ) ^ { 2 } - 4 \times 1 \times 56 } } { 2 \times 1 }$

$x = \dfrac { 30 \pm \sqrt{ \left ( \color{#FF6800}{ - } \color{#FF6800}{ 30 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 56 } } { 2 \times 1 }$

$ $ Remove negative signs because negative numbers raised to even powers are positive $ $

$x = \dfrac { 30 \pm \sqrt{ 30 ^ { 2 } - 4 \times 1 \times 56 } } { 2 \times 1 }$

$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 30 } \pm \sqrt{ \color{#FF6800}{ 30 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \color{#FF6800}{ 56 } } } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } } }$

$ $ Organize the expression $ $

$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 30 } \pm \sqrt{ \color{#FF6800}{ 676 } } } { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } } }$

$x = \dfrac { 30 \pm \sqrt{ \color{#FF6800}{ 676 } } } { 2 \times 1 }$

$ $ Organize the part that can be taken out of the radical sign inside the square root symbol $ $

$x = \dfrac { 30 \pm \color{#FF6800}{ 26 } } { 2 \times 1 }$

$x = \dfrac { 30 \pm 26 } { 2 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } }$

$ $ Multiplying any number by 1 does not change the value $ $

$x = \dfrac { 30 \pm 26 } { \color{#FF6800}{ 2 } }$

$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 30 } \pm \color{#FF6800}{ 26 } } { \color{#FF6800}{ 2 } } }$

$ $ Separate the answer $ $

$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 30 } \color{#FF6800}{ + } \color{#FF6800}{ 26 } } { \color{#FF6800}{ 2 } } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 30 } \color{#FF6800}{ - } \color{#FF6800}{ 26 } } { \color{#FF6800}{ 2 } } } \end{array}$

$\begin{array} {l} x = \dfrac { \color{#FF6800}{ 30 } \color{#FF6800}{ + } \color{#FF6800}{ 26 } } { 2 } \\ x = \dfrac { 30 - 26 } { 2 } \end{array}$

$ $ Add $ 30 $ and $ 26$

$\begin{array} {l} x = \dfrac { \color{#FF6800}{ 56 } } { 2 } \\ x = \dfrac { 30 - 26 } { 2 } \end{array}$

$\begin{array} {l} x = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 56 } } { \color{#FF6800}{ 2 } } } \\ x = \dfrac { 30 - 26 } { 2 } \end{array}$

$ $ Do the reduction of the fraction format $ $

$\begin{array} {l} x = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 28 } } { \color{#FF6800}{ 1 } } } \\ x = \dfrac { 30 - 26 } { 2 } \end{array}$

$\begin{array} {l} x = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 28 } } { \color{#FF6800}{ 1 } } } \\ x = \dfrac { 30 - 26 } { 2 } \end{array}$

$ $ Reduce the fraction to the lowest term $ $

$\begin{array} {l} x = \color{#FF6800}{ 28 } \\ x = \dfrac { 30 - 26 } { 2 } \end{array}$

$\begin{array} {l} x = 28 \\ x = \dfrac { \color{#FF6800}{ 30 } \color{#FF6800}{ - } \color{#FF6800}{ 26 } } { 2 } \end{array}$

$ $ Subtract $ 26 $ from $ 30$

$\begin{array} {l} x = 28 \\ x = \dfrac { \color{#FF6800}{ 4 } } { 2 } \end{array}$

$\begin{array} {l} x = 28 \\ x = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 4 } } { \color{#FF6800}{ 2 } } } \end{array}$

$ $ Do the reduction of the fraction format $ $

$\begin{array} {l} x = 28 \\ x = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 2 } } { \color{#FF6800}{ 1 } } } \end{array}$

$\begin{array} {l} x = 28 \\ x = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 2 } } { \color{#FF6800}{ 1 } } } \end{array}$

$ $ Reduce the fraction to the lowest term $ $

$\begin{array} {l} x = 28 \\ x = \color{#FF6800}{ 2 } \end{array}$

$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 30 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 56 } = \color{#FF6800}{ 0 }$

$ $ Determine the number of roots using discriminant, $ D=b^{2}-4ac $ from quadratic equation, $ ax^{2}+bx+c=0$

$\color{#FF6800}{ D } = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 30 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \color{#FF6800}{ 56 }$

$D = \left ( \color{#FF6800}{ - } \color{#FF6800}{ 30 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 56$

$ $ Remove negative signs because negative numbers raised to even powers are positive $ $

$D = 30 ^ { 2 } - 4 \times 1 \times 56$

$D = \color{#FF6800}{ 30 } ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times 56$

$ $ Calculate power $ $

$D = \color{#FF6800}{ 900 } - 4 \times 1 \times 56$

$D = 900 - 4 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \times 56$

$ $ Multiplying any number by 1 does not change the value $ $

$D = 900 - 4 \times 56$

$D = 900 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 56 }$

$ $ Multiply $ - 4 $ and $ 56$

$D = 900 \color{#FF6800}{ - } \color{#FF6800}{ 224 }$

$D = \color{#FF6800}{ 900 } \color{#FF6800}{ - } \color{#FF6800}{ 224 }$

$ $ Subtract $ 224 $ from $ 900$

$D = \color{#FF6800}{ 676 }$

$\color{#FF6800}{ D } = \color{#FF6800}{ 676 }$

$ $ Since $ D>0 $ , the number of real root of the following quadratic equation is 2 $ $

$ $ 2 real roots $ $

$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 30 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 56 } = \color{#FF6800}{ 0 }$

$ $ In the quadratic equation $ ax^{2}+bx+c=0 $ , if the two roots are $ \alpha, \beta $ , then it is $ \alpha + \beta =-\dfrac{b}{a} $ , $ \alpha\times\beta=\dfrac{c}{a}$

$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 30 } } { \color{#FF6800}{ 1 } } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 56 } } { \color{#FF6800}{ 1 } } }$

$\alpha + \beta = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 30 } } { \color{#FF6800}{ 1 } } } , \alpha \beta = \dfrac { 56 } { 1 }$

$ $ Solve the sign of a fraction with a negative sign $ $

$\alpha + \beta = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 30 } } { \color{#FF6800}{ 1 } } } , \alpha \beta = \dfrac { 56 } { 1 }$

$\alpha + \beta = \dfrac { 30 } { \color{#FF6800}{ 1 } } , \alpha \beta = \dfrac { 56 } { 1 }$

$ $ If the denominator is 1, the denominator can be removed $ $

$\alpha + \beta = \color{#FF6800}{ 30 } , \alpha \beta = \dfrac { 56 } { 1 }$

$\alpha + \beta = 30 , \alpha \beta = \dfrac { 56 } { \color{#FF6800}{ 1 } }$

$ $ If the denominator is 1, the denominator can be removed $ $

$\alpha + \beta = 30 , \alpha \beta = \color{#FF6800}{ 56 }$