Solve Using the Quadratic Formula x^2+4x+4=25

$x^{2} + 4 x + 4 = 25$

Step 1

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

Subtract $25$ from both sides of the equation.

$x^{2} + 4 x + 4 - 25 = 0$

Subtract $25$ from $4$ .

$x^{2} + 4 x - 21 = 0$

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 $a = 1$ , $b = 4$ , and $c = - 21$ into the quadratic formula and solve for $x$ .

$\frac{- 4 \pm \sqrt{4^{2} - 4 \cdot (1 \cdot - 21)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $4$ to the power of $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot - 21}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 21$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot - 21}}{2 \cdot 1}$

Multiply $- 4$ by $- 21$ .

$x = \frac{- 4 \pm \sqrt{16 + 84}}{2 \cdot 1}$

Add $16$ and $84$ .

$x = \frac{- 4 \pm \sqrt{100}}{2 \cdot 1}$

Rewrite $100$ as $10^{2}$ .

$x = \frac{- 4 \pm \sqrt{10^{2}}}{2 \cdot 1}$

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

$x = \frac{- 4 \pm 10}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 4 \pm 10}{2}$

Simplify $\frac{- 4 \pm 10}{2}$ .

$x = - 2 \pm 5$

Step 5

The final answer is the combination of both solutions.

$x = 3, - 7$