Solve Using the Quadratic Formula 4x^2-6x-10=0

$4 x^{2} - 6 x - 10 = 0$

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 $a = 4$ , $b = - 6$ , and $c = - 10$ into the quadratic formula and solve for $x$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (4 \cdot - 10)}}{2 \cdot 4}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 6$ to the power of $2$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 4 \cdot - 10}}{2 \cdot 4}$

Multiply $- 4 \cdot 4 \cdot - 10$ .

Multiply $- 4$ by $4$ .

$x = \frac{6 \pm \sqrt{36 - 16 \cdot - 10}}{2 \cdot 4}$

Multiply $- 16$ by $- 10$ .

$x = \frac{6 \pm \sqrt{36 + 160}}{2 \cdot 4}$

Add $36$ and $160$ .

$x = \frac{6 \pm \sqrt{196}}{2 \cdot 4}$

Rewrite $196$ as $14^{2}$ .

$x = \frac{6 \pm \sqrt{14^{2}}}{2 \cdot 4}$

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

$x = \frac{6 \pm 14}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{6 \pm 14}{8}$

Simplify $\frac{6 \pm 14}{8}$ .

$x = \frac{3 \pm 7}{4}$

Step 4

The final answer is the combination of both solutions.

$x = \frac{5}{2}, - 1$