Solve Using the Quadratic Formula 4x^2-13x-35=0

$4 x^{2} - 13 x - 35 = 0$

Step 1

Use the quadratic formula to find the solutions.

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

Step 2

Substitute the values $a = 4$ , $b = - 13$ , and $c = - 35$ into the quadratic formula and solve for $x$ .

$\frac{13 \pm \sqrt{{(- 13)}^{2} - 4 \cdot (4 \cdot - 35)}}{2 \cdot 4}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 13$ to the power of $2$ .

$x = \frac{13 \pm \sqrt{169 - 4 \cdot 4 \cdot - 35}}{2 \cdot 4}$

Multiply $- 4 \cdot 4 \cdot - 35$ .

Multiply $- 4$ by $4$ .

$x = \frac{13 \pm \sqrt{169 - 16 \cdot - 35}}{2 \cdot 4}$

Multiply $- 16$ by $- 35$ .

$x = \frac{13 \pm \sqrt{169 + 560}}{2 \cdot 4}$

Add $169$ and $560$ .

$x = \frac{13 \pm \sqrt{729}}{2 \cdot 4}$

Rewrite $729$ as $27^{2}$ .

$x = \frac{13 \pm \sqrt{27^{2}}}{2 \cdot 4}$

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

$x = \frac{13 \pm 27}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{13 \pm 27}{8}$

Step 4

The final answer is the combination of both solutions.

$x = 5, - \frac{7}{4}$