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

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

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 $c = 2$ into the quadratic formula and solve for $x$ .

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

Step 3

Simplify.

Simplify the numerator.

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

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

Multiply $- 4 \cdot 4 \cdot 2$ .

Multiply $- 4$ by $4$ .

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

Multiply $- 16$ by $2$ .

$x = \frac{- 6 \pm \sqrt{36 - 32}}{2 \cdot 4}$

Subtract $32$ from $36$ .

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

Rewrite $4$ as $2^{2}$ .

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

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

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

Multiply $2$ by $4$ .

$x = \frac{- 6 \pm 2}{8}$

Simplify $\frac{- 6 \pm 2}{8}$ .

$x = \frac{- 3 \pm 1}{4}$

Step 4

The final answer is the combination of both solutions.

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