Solve Using the Quadratic Formula 4x^2+20x+25=49

$4 x^{2} + 20 x + 25 = 49$

Step 1

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

Subtract $49$ from both sides of the equation.

$4 x^{2} + 20 x + 25 - 49 = 0$

Subtract $49$ from $25$ .

$4 x^{2} + 20 x - 24 = 0$

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 $a = 4$ , $b = 20$ , and $c = - 24$ into the quadratic formula and solve for $x$ .

$\frac{- 20 \pm \sqrt{20^{2} - 4 \cdot (4 \cdot - 24)}}{2 \cdot 4}$

Step 4

Simplify.

Simplify the numerator.

Raise $20$ to the power of $2$ .

$x = \frac{- 20 \pm \sqrt{400 - 4 \cdot 4 \cdot - 24}}{2 \cdot 4}$

Multiply $- 4 \cdot 4 \cdot - 24$ .

Multiply $- 4$ by $4$ .

$x = \frac{- 20 \pm \sqrt{400 - 16 \cdot - 24}}{2 \cdot 4}$

Multiply $- 16$ by $- 24$ .

$x = \frac{- 20 \pm \sqrt{400 + 384}}{2 \cdot 4}$

Add $400$ and $384$ .

$x = \frac{- 20 \pm \sqrt{784}}{2 \cdot 4}$

Rewrite $784$ as $28^{2}$ .

$x = \frac{- 20 \pm \sqrt{28^{2}}}{2 \cdot 4}$

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

$x = \frac{- 20 \pm 28}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{- 20 \pm 28}{8}$

Simplify $\frac{- 20 \pm 28}{8}$ .

$x = \frac{- 5 \pm 7}{2}$

Step 5

The final answer is the combination of both solutions.

$x = 1, - 6$