Solve Using the Quadratic Formula 4x^2-3x+9=2x+1

$4 x^{2} - 3 x + 9 = 2 x + 1$

Step 1

Move all terms to the left side of the equation and simplify.

Move all the expressions to the left side of the equation.

Subtract $2 x$ from both sides of the equation.

$4 x^{2} - 3 x + 9 - 2 x = 1$

Subtract $1$ from both sides of the equation.

$4 x^{2} - 3 x + 9 - 2 x - 1 = 0$

Simplify $4 x^{2} - 3 x + 9 - 2 x - 1$ .

Subtract $2 x$ from $- 3 x$ .

$4 x^{2} - 5 x + 9 - 1 = 0$

Subtract $1$ from $9$ .

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

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

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

Step 4

Simplify.

Simplify the numerator.

Raise $- 5$ to the power of $2$ .

$x = \frac{5 \pm \sqrt{25 - 4 \cdot 4 \cdot 8}}{2 \cdot 4}$

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

Multiply $- 4$ by $4$ .

$x = \frac{5 \pm \sqrt{25 - 16 \cdot 8}}{2 \cdot 4}$

Multiply $- 16$ by $8$ .

$x = \frac{5 \pm \sqrt{25 - 128}}{2 \cdot 4}$

Subtract $128$ from $25$ .

$x = \frac{5 \pm \sqrt{- 103}}{2 \cdot 4}$

Rewrite $- 103$ as $- 1 (103)$ .

$x = \frac{5 \pm \sqrt{- 1 \cdot 103}}{2 \cdot 4}$

Rewrite $\sqrt{- 1 (103)}$ as $\sqrt{- 1} \cdot \sqrt{103}$ .

$x = \frac{5 \pm \sqrt{- 1} \cdot \sqrt{103}}{2 \cdot 4}$

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

$x = \frac{5 \pm i \sqrt{103}}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{5 \pm i \sqrt{103}}{8}$