Solve Using the Quadratic Formula 4x^2+7=9x

$4 x^{2} + 7 = 9 x$

Step 1

Subtract $9 x$ from both sides of the equation.

$4 x^{2} + 7 - 9 x = 0$

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 $a = 4$ , $b = - 9$ , and $c = 7$ into the quadratic formula and solve for $x$ .

$\frac{9 \pm \sqrt{{(- 9)}^{2} - 4 \cdot (4 \cdot 7)}}{2 \cdot 4}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 9$ to the power of $2$ .

$x = \frac{9 \pm \sqrt{81 - 4 \cdot 4 \cdot 7}}{2 \cdot 4}$

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

Multiply $- 4$ by $4$ .

$x = \frac{9 \pm \sqrt{81 - 16 \cdot 7}}{2 \cdot 4}$

Multiply $- 16$ by $7$ .

$x = \frac{9 \pm \sqrt{81 - 112}}{2 \cdot 4}$

Subtract $112$ from $81$ .

$x = \frac{9 \pm \sqrt{- 31}}{2 \cdot 4}$

Rewrite $- 31$ as $- 1 (31)$ .

$x = \frac{9 \pm \sqrt{- 1 \cdot 31}}{2 \cdot 4}$

Rewrite $\sqrt{- 1 (31)}$ as $\sqrt{- 1} \cdot \sqrt{31}$ .

$x = \frac{9 \pm \sqrt{- 1} \cdot \sqrt{31}}{2 \cdot 4}$

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

$x = \frac{9 \pm i \sqrt{31}}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{9 \pm i \sqrt{31}}{8}$