# Evaluate: -5w^2+180=0

## Expression: $-5{w}^{2}+180=0$

Divide both sides of the equation by $-5$

${w}^{2}-36=0$

Identify the coefficients $a$, $b$ and $c$ of the quadratic equation

$\begin{array} { l }a=1,& b=0,& c=-36\end{array}$

Substitute $a=1$, $b=0$ and $c=-36$ into the quadratic formula $w=\frac{ -b\pm\sqrt{ {b}^{2}-4ac } }{ 2a }$

$w=\frac{ -0\pm\sqrt{ {0}^{2}-4 \times 1 \times \left( -36 \right) } }{ 2 \times 1 }$

Removing $0$ doesn't change the value, so remove it from the expression

$w=\frac{ \sqrt{ {0}^{2}-4 \times 1 \times \left( -36 \right) } }{ 2 \times 1 }$

Any expression multiplied by $1$ remains the same

$w=\frac{ \sqrt{ {0}^{2}-4 \times \left( -36 \right) } }{ 2 \times 1 }$

Any expression multiplied by $1$ remains the same

$w=\frac{ \sqrt{ {0}^{2}-4 \times \left( -36 \right) } }{ 2 }$

$0$ raised to any positive power equals $0$

$w=\frac{ \sqrt{ 0-4 \times \left( -36 \right) } }{ 2 }$

Multiply the numbers

$w=\frac{ \sqrt{ 0+144 } }{ 2 }$

Removing $0$ doesn't change the value, so remove it from the expression

$w=\frac{ \sqrt{ 144 } }{ 2 }$

Evaluate the square root

$w=\frac{ 12 }{ 2 }$

Write the solutions, one with a $+$ sign and one with a $-$ sign

$\begin{array} { l }w=\frac{ 12 }{ 2 },\\w=\frac{ -12 }{ 2 }\end{array}$

Cancel out the common factor $2$

$\begin{array} { l }w=6,\\w=\frac{ -12 }{ 2 }\end{array}$

Cancel out the common factor $2$

$\begin{array} { l }w=6,\\w=-6\end{array}$

The equation has $2$ solutions

$\begin{array} { l }w_1=-6,& w_2=6\end{array}$

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