# Solve for: 2x^2=128

## Expression: $2{x}^{2}=128$

Divide both sides of the equation by $2$

${x}^{2}=64$

Move the constant to the left-hand side and change its sign

${x}^{2}-64=0$

Use ${a}^{2}-{b}^{2}=\left( a-b \right)\left( a+b \right)$ to factor the expression

$\left( x-8 \right) \times \left( x+8 \right)=0$

When the product of factors equals $0$, at least one factor is $0$

$\begin{array} { l }x-8=0,\\x+8=0\end{array}$

Solve the equation for $x$

$\begin{array} { l }x=8,\\x+8=0\end{array}$

Solve the equation for $x$

$\begin{array} { l }x=8,\\x=-8\end{array}$

The equation has $2$ solutions

$\begin{array} { l }x_1=-8,& x_2=8\end{array}$

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