Solve Using the Quadratic Formula 16x^2+1=5x

$16 x^{2} + 1 = 5 x$

Step 1

Subtract $5 x$ from both sides of the equation.

$16 x^{2} + 1 - 5 x = 0$

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 $a = 16$ , $b = - 5$ , and $c = 1$ into the quadratic formula and solve for $x$ .

$\frac{5 \pm \sqrt{{(- 5)}^{2} - 4 \cdot (16 \cdot 1)}}{2 \cdot 16}$

Step 4

Simplify.

Simplify the numerator.

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

$x = \frac{5 \pm \sqrt{25 - 4 \cdot 16 \cdot 1}}{2 \cdot 16}$

Multiply $- 4 \cdot 16 \cdot 1$ .

Multiply $- 4$ by $16$ .

$x = \frac{5 \pm \sqrt{25 - 64 \cdot 1}}{2 \cdot 16}$

Multiply $- 64$ by $1$ .

$x = \frac{5 \pm \sqrt{25 - 64}}{2 \cdot 16}$

Subtract $64$ from $25$ .

$x = \frac{5 \pm \sqrt{- 39}}{2 \cdot 16}$

Rewrite $- 39$ as $- 1 (39)$ .

$x = \frac{5 \pm \sqrt{- 1 \cdot 39}}{2 \cdot 16}$

Rewrite $\sqrt{- 1 (39)}$ as $\sqrt{- 1} \cdot \sqrt{39}$ .

$x = \frac{5 \pm \sqrt{- 1} \cdot \sqrt{39}}{2 \cdot 16}$

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

$x = \frac{5 \pm i \sqrt{39}}{2 \cdot 16}$

Multiply $2$ by $16$ .

$x = \frac{5 \pm i \sqrt{39}}{32}$