Find the Roots (Zeros) y=8x^2+4x+5

$y = 8 x^{2} + 4 x + 5$

Step 1

Set $8 x^{2} + 4 x + 5$ equal to $0$ .

$8 x^{2} + 4 x + 5 = 0$

Step 2

Solve for $x$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 8$ , $b = 4$ , and $c = 5$ into the quadratic formula and solve for $x$ .

$\frac{- 4 \pm \sqrt{4^{2} - 4 \cdot (8 \cdot 5)}}{2 \cdot 8}$

Simplify.

Simplify the numerator.

Raise $4$ to the power of $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 8 \cdot 5}}{2 \cdot 8}$

Multiply $- 4 \cdot 8 \cdot 5$ .

Multiply $- 4$ by $8$ .

$x = \frac{- 4 \pm \sqrt{16 - 32 \cdot 5}}{2 \cdot 8}$

Multiply $- 32$ by $5$ .

$x = \frac{- 4 \pm \sqrt{16 - 160}}{2 \cdot 8}$

Subtract $160$ from $16$ .

$x = \frac{- 4 \pm \sqrt{- 144}}{2 \cdot 8}$

Rewrite $- 144$ as $- 1 (144)$ .

$x = \frac{- 4 \pm \sqrt{- 1 \cdot 144}}{2 \cdot 8}$

Rewrite $\sqrt{- 1 (144)}$ as $\sqrt{- 1} \cdot \sqrt{144}$ .

$x = \frac{- 4 \pm \sqrt{- 1} \cdot \sqrt{144}}{2 \cdot 8}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 4 \pm i \cdot \sqrt{144}}{2 \cdot 8}$

Rewrite $144$ as $12^{2}$ .

$x = \frac{- 4 \pm i \cdot \sqrt{12^{2}}}{2 \cdot 8}$

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

$x = \frac{- 4 \pm i \cdot 12}{2 \cdot 8}$

Move $12$ to the left of $i$ .

$x = \frac{- 4 \pm 12 i}{2 \cdot 8}$

Multiply $2$ by $8$ .

$x = \frac{- 4 \pm 12 i}{16}$

Simplify $\frac{- 4 \pm 12 i}{16}$ .

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

The final answer is the combination of both solutions.

$x = - \frac{1}{4} + \frac{3 i}{4}, - \frac{1}{4} - \frac{3 i}{4}$

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

Step 3