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Formula
Solve the quadratic equation
Answer
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$$x ^ { 2 } - 2 x - 15 = 0$$
$\begin{array} {l} x = 5 \\ x = - 3 \end{array}$
Calculate using the quadratic formula
$x = \dfrac { \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } 2 \right ) \pm \sqrt{ \left ( - 2 \right ) ^ { 2 } - 4 \times 1 \times \left ( - 15 \right ) } } { 2 \times 1 }$
$ $ Simplify Minus $ $
$x = \dfrac { 2 \pm \sqrt{ \left ( - 2 \right ) ^ { 2 } - 4 \times 1 \times \left ( - 15 \right ) } } { 2 \times 1 }$
$x = \dfrac { 2 \pm \sqrt{ \left ( \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times \left ( - 15 \right ) } } { 2 \times 1 }$
$ $ Remove negative signs because negative numbers raised to even powers are positive $ $
$x = \dfrac { 2 \pm \sqrt{ 2 ^ { 2 } - 4 \times 1 \times \left ( - 15 \right ) } } { 2 \times 1 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 2 \pm \sqrt{ 2 ^ { 2 } - 4 \times 1 \times \left ( - 15 \right ) } } { 2 \times 1 } }$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 2 \pm \sqrt{ 64 } } { 2 \times 1 } }$
$x = \dfrac { 2 \pm \sqrt{ \color{#FF6800}{ 64 } } } { 2 \times 1 }$
$ $ Organize the part that can be taken out of the radical sign inside the square root symbol $ $
$x = \dfrac { 2 \pm \color{#FF6800}{ 8 } } { 2 \times 1 }$
$x = \dfrac { 2 \pm 8 } { 2 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } }$
$ $ Multiplying any number by 1 does not change the value $ $
$x = \dfrac { 2 \pm 8 } { \color{#FF6800}{ 2 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 2 \pm 8 } { 2 } }$
$ $ Separate the answer $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 2 + 8 } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 2 - 8 } { 2 } } \end{array}$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ 8 } } { 2 } \\ x = \dfrac { 2 - 8 } { 2 } \end{array}$
$ $ Add $ 2 $ and $ 8$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ 10 } } { 2 } \\ x = \dfrac { 2 - 8 } { 2 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 10 } { 2 } } \\ x = \dfrac { 2 - 8 } { 2 } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 5 } { 1 } } \\ x = \dfrac { 2 - 8 } { 2 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 5 } { 1 } } \\ x = \dfrac { 2 - 8 } { 2 } \end{array}$
$ $ Reduce the fraction to the lowest term $ $
$\begin{array} {l} x = \color{#FF6800}{ 5 } \\ x = \dfrac { 2 - 8 } { 2 } \end{array}$
$\begin{array} {l} x = 5 \\ x = \dfrac { \color{#FF6800}{ 2 } \color{#FF6800}{ - } \color{#FF6800}{ 8 } } { 2 } \end{array}$
$ $ Subtract $ 8 $ from $ 2$
$\begin{array} {l} x = 5 \\ x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 6 } } { 2 } \end{array}$
$\begin{array} {l} x = 5 \\ x = \color{#FF6800}{ \dfrac { - 6 } { 2 } } \end{array}$
$ $ Do the reduction of the fraction format $ $
$\begin{array} {l} x = 5 \\ x = \color{#FF6800}{ \dfrac { - 3 } { 1 } } \end{array}$
$\begin{array} {l} x = 5 \\ x = \dfrac { - 3 } { \color{#FF6800}{ 1 } } \end{array}$
$ $ If the denominator is 1, the denominator can be removed $ $
$\begin{array} {l} x = 5 \\ x = \color{#FF6800}{ - } \color{#FF6800}{ 3 } \end{array}$
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