Solve Using the Quadratic Formula x^2+4x+20=12x-5

$x^{2} + 4 x + 20 = 12 x - 5$

Step 1

Move all terms to the left side of the equation and simplify.

Move all the expressions to the left side of the equation.

Subtract $12 x$ from both sides of the equation.

$x^{2} + 4 x + 20 - 12 x = - 5$

Add $5$ to both sides of the equation.

$x^{2} + 4 x + 20 - 12 x + 5 = 0$

Simplify $x^{2} + 4 x + 20 - 12 x + 5$ .

Subtract $12 x$ from $4 x$ .

$x^{2} - 8 x + 20 + 5 = 0$

Add $20$ and $5$ .

$x^{2} - 8 x + 25 = 0$

Step 2

Use the quadratic formula to find the solutions.

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

Step 3

Substitute the values $a = 1$ , $b = - 8$ , and $c = 25$ into the quadratic formula and solve for $x$ .

$\frac{8 \pm \sqrt{{(- 8)}^{2} - 4 \cdot (1 \cdot 25)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 8$ to the power of $2$ .

$x = \frac{8 \pm \sqrt{64 - 4 \cdot 1 \cdot 25}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 25$ .

Multiply $- 4$ by $1$ .

$x = \frac{8 \pm \sqrt{64 - 4 \cdot 25}}{2 \cdot 1}$

Multiply $- 4$ by $25$ .

$x = \frac{8 \pm \sqrt{64 - 100}}{2 \cdot 1}$

Subtract $100$ from $64$ .

$x = \frac{8 \pm \sqrt{- 36}}{2 \cdot 1}$

Rewrite $- 36$ as $- 1 (36)$ .

$x = \frac{8 \pm \sqrt{- 1 \cdot 36}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (36)}$ as $\sqrt{- 1} \cdot \sqrt{36}$ .

$x = \frac{8 \pm \sqrt{- 1} \cdot \sqrt{36}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{8 \pm i \cdot \sqrt{36}}{2 \cdot 1}$

Rewrite $36$ as $6^{2}$ .

$x = \frac{8 \pm i \cdot \sqrt{6^{2}}}{2 \cdot 1}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{8 \pm i \cdot 6}{2 \cdot 1}$

Move $6$ to the left of $i$ .

$x = \frac{8 \pm 6 i}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{8 \pm 6 i}{2}$

Simplify $\frac{8 \pm 6 i}{2}$ .

$x = 4 \pm 3 i$

Solve Using the Quadratic Formula x^2+4x+20=12x-5

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