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

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

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 $c = - 4$ into the quadratic formula and solve for $x$ .

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

Step 3

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $4$ .

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

Multiply $- 16$ by $- 4$ .

$x = \frac{- 6 \pm \sqrt{36 + 64}}{2 \cdot 4}$

Add $36$ and $64$ .

$x = \frac{- 6 \pm \sqrt{100}}{2 \cdot 4}$

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

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

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

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

Multiply $2$ by $4$ .

$x = \frac{- 6 \pm 10}{8}$

Simplify $\frac{- 6 \pm 10}{8}$ .

$x = \frac{- 3 \pm 5}{4}$

Step 4

The final answer is the combination of both solutions.

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