Solve for x x^2+3x+7=-|2x|+12

$x^{2} + 3 x + 7 = - | 2 x | + 12$

Rewrite the equation as $- | 2 x | + 12 = x^{2} + 3 x + 7$ .

$- | 2 x | + 12 = x^{2} + 3 x + 7$

Move all terms not containing $| 2 x |$ to the right side of the equation.

Subtract $12$ from both sides of the equation.

$- | 2 x | = x^{2} + 3 x + 7 - 12$

Subtract $12$ from $7$ .

$- | 2 x | = x^{2} + 3 x - 5$

Divide each term in $- | 2 x | = x^{2} + 3 x - 5$ by $- 1$ and simplify.

Divide each term in $- | 2 x | = x^{2} + 3 x - 5$ by $- 1$ .

$\frac{- | 2 x |}{- 1} = \frac{x^{2}}{- 1} + \frac{3 x}{- 1} + \frac{- 5}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{| 2 x |}{1} = \frac{x^{2}}{- 1} + \frac{3 x}{- 1} + \frac{- 5}{- 1}$

Divide $| 2 x |$ by $1$ .

$| 2 x | = \frac{x^{2}}{- 1} + \frac{3 x}{- 1} + \frac{- 5}{- 1}$

Simplify the right side.

Simplify each term.

Move the negative one from the denominator of $\frac{x^{2}}{- 1}$ .

$| 2 x | = - 1 \cdot x^{2} + \frac{3 x}{- 1} + \frac{- 5}{- 1}$

Rewrite $- 1 \cdot x^{2}$ as $- x^{2}$ .

$| 2 x | = - x^{2} + \frac{3 x}{- 1} + \frac{- 5}{- 1}$

Move the negative one from the denominator of $\frac{3 x}{- 1}$ .

$| 2 x | = - x^{2} - 1 \cdot (3 x) + \frac{- 5}{- 1}$

Rewrite $- 1 \cdot (3 x)$ as $- (3 x)$ .

$| 2 x | = - x^{2} - (3 x) + \frac{- 5}{- 1}$

Multiply $3$ by $- 1$ .

$| 2 x | = - x^{2} - 3 x + \frac{- 5}{- 1}$

Divide $- 5$ by $- 1$ .

$| 2 x | = - x^{2} - 3 x + 5$

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$2 x = \pm (- x^{2} - 3 x + 5)$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$2 x = - x^{2} - 3 x + 5$

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$- x^{2} - 3 x + 5 = 2 x$

Move all terms containing $x$ to the left side of the equation.

Subtract $2 x$ from both sides of the equation.

$- x^{2} - 3 x + 5 - 2 x = 0$

Subtract $2 x$ from $- 3 x$ .

$- x^{2} - 5 x + 5 = 0$

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = - 1$ , $b = - 5$ , and $c = 5$ into the quadratic formula and solve for $x$ .

$\frac{5 \pm \sqrt{{(- 5)}^{2} - 4 \cdot (- 1 \cdot 5)}}{2 \cdot - 1}$

Simplify.

Simplify the numerator.

Raise $- 5$ to the power of $2$ .

$x = \frac{5 \pm \sqrt{25 - 4 \cdot - 1 \cdot 5}}{2 \cdot - 1}$

Multiply $- 4$ by $- 1$ .

$x = \frac{5 \pm \sqrt{25 + 4 \cdot 5}}{2 \cdot - 1}$

Multiply $4$ by $5$ .

$x = \frac{5 \pm \sqrt{25 + 20}}{2 \cdot - 1}$

Add $25$ and $20$ .

$x = \frac{5 \pm \sqrt{45}}{2 \cdot - 1}$

Rewrite $45$ as $3^{2} \cdot 5$ .

Factor $9$ out of $45$ .

$x = \frac{5 \pm \sqrt{9 (5)}}{2 \cdot - 1}$

Rewrite $9$ as $3^{2}$ .

$x = \frac{5 \pm \sqrt{3^{2} \cdot 5}}{2 \cdot - 1}$

Pull terms out from under the radical.

$x = \frac{5 \pm 3 \sqrt{5}}{2 \cdot - 1}$

Multiply $2$ by $- 1$ .

$x = \frac{5 \pm 3 \sqrt{5}}{- 2}$

Move the negative in front of the fraction.

$x = - \frac{5 \pm 3 \sqrt{5}}{2}$

The final answer is the combination of both solutions.

$x = - \frac{5 + 3 \sqrt{5}}{2}, - \frac{5 - 3 \sqrt{5}}{2}$

Next, use the negative value of the $\pm$ to find the second solution.

$2 x = - (- x^{2} - 3 x + 5)$

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$- (- x^{2} - 3 x + 5) = 2 x$

Simplify $- (- x^{2} - 3 x + 5)$ .

Start simplifying.

$0 + 0 - (- x^{2} - 3 x + 5) = 2 x$

Simplify by adding zeros.

$- (- x^{2} - 3 x + 5) = 2 x$

Apply the distributive property.

$- - x^{2} - (- 3 x) - 1 \cdot 5 = 2 x$

Simplify.

Multiply $- - x^{2}$ .

Multiply $- 1$ by $- 1$ .

$1 x^{2} - (- 3 x) - 1 \cdot 5 = 2 x$

Multiply $x^{2}$ by $1$ .

$x^{2} - (- 3 x) - 1 \cdot 5 = 2 x$

Multiply $- 3$ by $- 1$ .

$x^{2} + 3 x - 1 \cdot 5 = 2 x$

Multiply $- 1$ by $5$ .

$x^{2} + 3 x - 5 = 2 x$

Move all terms containing $x$ to the left side of the equation.

Subtract $2 x$ from both sides of the equation.

$x^{2} + 3 x - 5 - 2 x = 0$

Subtract $2 x$ from $3 x$ .

$x^{2} + x - 5 = 0$

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = 1$ , and $c = - 5$ into the quadratic formula and solve for $x$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (1 \cdot - 5)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1 \cdot - 5}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot - 5}}{2 \cdot 1}$

Multiply $- 4$ by $- 5$ .

$x = \frac{- 1 \pm \sqrt{1 + 20}}{2 \cdot 1}$

Add $1$ and $20$ .

$x = \frac{- 1 \pm \sqrt{21}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 1 \pm \sqrt{21}}{2}$

The final answer is the combination of both solutions.

$x = - \frac{1 - \sqrt{21}}{2}, - \frac{1 + \sqrt{21}}{2}$

The complete solution is the result of both the positive and negative portions of the solution.

$x = - \frac{5 + 3 \sqrt{5}}{2}, - \frac{5 - 3 \sqrt{5}}{2}, - \frac{1 - \sqrt{21}}{2}, - \frac{1 + \sqrt{21}}{2}$

Exclude the solutions that do not make $x^{2} + 3 x + 7 = - | 2 x | + 12$ true.

$x = - \frac{5 - 3 \sqrt{5}}{2}, - \frac{1 + \sqrt{21}}{2}$

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{5 - 3 \sqrt{5}}{2}, - \frac{1 + \sqrt{21}}{2}$

Decimal Form:

$x = 0.85410196 \dots, - 2.79128784 \dots$

Solve for x x^2+3x+7=-|2x|+12

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