Solve Using the Quadratic Formula 8x^2+1=3x

$8 x^{2} + 1 = 3 x$

Step 1

Subtract $3 x$ from both sides of the equation.

$8 x^{2} + 1 - 3 x = 0$

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 $a = 8$ , $b = - 3$ , and $c = 1$ into the quadratic formula and solve for $x$ .

$\frac{3 \pm \sqrt{{(- 3)}^{2} - 4 \cdot (8 \cdot 1)}}{2 \cdot 8}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 3$ to the power of $2$ .

$x = \frac{3 \pm \sqrt{9 - 4 \cdot 8 \cdot 1}}{2 \cdot 8}$

Multiply $- 4 \cdot 8 \cdot 1$ .

Multiply $- 4$ by $8$ .

$x = \frac{3 \pm \sqrt{9 - 32 \cdot 1}}{2 \cdot 8}$

Multiply $- 32$ by $1$ .

$x = \frac{3 \pm \sqrt{9 - 32}}{2 \cdot 8}$

Subtract $32$ from $9$ .

$x = \frac{3 \pm \sqrt{- 23}}{2 \cdot 8}$

Rewrite $- 23$ as $- 1 (23)$ .

$x = \frac{3 \pm \sqrt{- 1 \cdot 23}}{2 \cdot 8}$

Rewrite $\sqrt{- 1 (23)}$ as $\sqrt{- 1} \cdot \sqrt{23}$ .

$x = \frac{3 \pm \sqrt{- 1} \cdot \sqrt{23}}{2 \cdot 8}$

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

$x = \frac{3 \pm i \sqrt{23}}{2 \cdot 8}$

Multiply $2$ by $8$ .

$x = \frac{3 \pm i \sqrt{23}}{16}$