Find the Roots (Zeros) 7x^2+3=8x

$7 x^{2} + 3 = 8 x$

Step 1

Subtract $8 x$ from both sides of the equation.

$7 x^{2} + 3 - 8 x = 0$

Step 2

Use the quadratic formula to find the solutions.

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

Step 3

Substitute the values $a = 7$ , $b = - 8$ , and $c = 3$ into the quadratic formula and solve for $x$ .

$\frac{8 \pm \sqrt{{(- 8)}^{2} - 4 \cdot (7 \cdot 3)}}{2 \cdot 7}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 8$ to the power of $2$ .

$x = \frac{8 \pm \sqrt{64 - 4 \cdot 7 \cdot 3}}{2 \cdot 7}$

Multiply $- 4 \cdot 7 \cdot 3$ .

Multiply $- 4$ by $7$ .

$x = \frac{8 \pm \sqrt{64 - 28 \cdot 3}}{2 \cdot 7}$

Multiply $- 28$ by $3$ .

$x = \frac{8 \pm \sqrt{64 - 84}}{2 \cdot 7}$

Subtract $84$ from $64$ .

$x = \frac{8 \pm \sqrt{- 20}}{2 \cdot 7}$

Rewrite $- 20$ as $- 1 (20)$ .

$x = \frac{8 \pm \sqrt{- 1 \cdot 20}}{2 \cdot 7}$

Rewrite $\sqrt{- 1 (20)}$ as $\sqrt{- 1} \cdot \sqrt{20}$ .

$x = \frac{8 \pm \sqrt{- 1} \cdot \sqrt{20}}{2 \cdot 7}$

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

$x = \frac{8 \pm i \cdot \sqrt{20}}{2 \cdot 7}$

Rewrite $20$ as $2^{2} \cdot 5$ .

Factor $4$ out of $20$ .

$x = \frac{8 \pm i \cdot \sqrt{4 (5)}}{2 \cdot 7}$

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

$x = \frac{8 \pm i \cdot \sqrt{2^{2} \cdot 5}}{2 \cdot 7}$

Pull terms out from under the radical.

$x = \frac{8 \pm i \cdot (2 \sqrt{5})}{2 \cdot 7}$

Move $2$ to the left of $i$ .

$x = \frac{8 \pm 2 i \sqrt{5}}{2 \cdot 7}$

Multiply $2$ by $7$ .

$x = \frac{8 \pm 2 i \sqrt{5}}{14}$

Simplify $\frac{8 \pm 2 i \sqrt{5}}{14}$ .

$x = \frac{4 \pm i \sqrt{5}}{7}$

Step 5