# Evaluate: |2x+3|^2=5 * |2x+3|+14

## Expression: ${|2x+3|}^{2}=5 \times |2x+3|+14$

Use ${|a|}^{2}={a}^{2}$ to simplify the expression

${\left( 2x+3 \right)}^{2}=5 \times |2x+3|+14$

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

$4{x}^{2}+12x+9=5 \times |2x+3|+14$

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

$4{x}^{2}+12x+9-5 \times |2x+3|=14$

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

$4{x}^{2}+12x-5 \times |2x+3|=14-9$

Subtract the numbers

$4{x}^{2}+12x-5 \times |2x+3|=5$

Separate the equation into $2$ possible cases

$\begin{array} { l }\begin{array} { l }4{x}^{2}+12x-5\left( 2x+3 \right)=5,& 2x+3 \geq 0\end{array},\\\begin{array} { l }4{x}^{2}+12x-5 \times \left( -\left( 2x+3 \right) \right)=5,& 2x+3 < 0\end{array}\end{array}$

Solve the equation for $x$

$\begin{array} { l }\begin{array} { l }\begin{array} { l }x=-\frac{ 5 }{ 2 },\\x=2\end{array},& 2x+3 \geq 0\end{array},\\\begin{array} { l }4{x}^{2}+12x-5 \times \left( -\left( 2x+3 \right) \right)=5,& 2x+3 < 0\end{array}\end{array}$

Solve the inequality for $x$

$\begin{array} { l }\begin{array} { l }\begin{array} { l }x=-\frac{ 5 }{ 2 },\\x=2\end{array},& x \geq -\frac{ 3 }{ 2 }\end{array},\\\begin{array} { l }4{x}^{2}+12x-5 \times \left( -\left( 2x+3 \right) \right)=5,& 2x+3 < 0\end{array}\end{array}$

Solve the equation for $x$

$\begin{array} { l }\begin{array} { l }\begin{array} { l }x=-\frac{ 5 }{ 2 },\\x=2\end{array},& x \geq -\frac{ 3 }{ 2 }\end{array},\\\begin{array} { l }\begin{array} { l }x=-5,\\x=-\frac{ 1 }{ 2 }\end{array},& 2x+3 < 0\end{array}\end{array}$

Solve the inequality for $x$

$\begin{array} { l }\begin{array} { l }\begin{array} { l }x=-\frac{ 5 }{ 2 },\\x=2\end{array},& x \geq -\frac{ 3 }{ 2 }\end{array},\\\begin{array} { l }\begin{array} { l }x=-5,\\x=-\frac{ 1 }{ 2 }\end{array},& x < -\frac{ 3 }{ 2 }\end{array}\end{array}$

Find the intersection

$\begin{array} { l }x=2,\\\begin{array} { l }\begin{array} { l }x=-5,\\x=-\frac{ 1 }{ 2 }\end{array},& x < -\frac{ 3 }{ 2 }\end{array}\end{array}$

Find the intersection

$\begin{array} { l }x=2,\\x=-5\end{array}$

The equation has $2$ solutions

$\begin{array} { l }x_1=-5,& x_2=2\end{array}$

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