Solve Using the Quadratic Formula 16x^2+2x-3=0

$16 x^{2} + 2 x - 3 = 0$

Step 1

Use the quadratic formula to find the solutions.

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

Step 2

Substitute the values $a = 16$ , $b = 2$ , and $c = - 3$ into the quadratic formula and solve for $x$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (16 \cdot - 3)}}{2 \cdot 16}$

Step 3

Simplify.

Simplify the numerator.

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

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 16 \cdot - 3}}{2 \cdot 16}$

Multiply $- 4 \cdot 16 \cdot - 3$ .

Multiply $- 4$ by $16$ .

$x = \frac{- 2 \pm \sqrt{4 - 64 \cdot - 3}}{2 \cdot 16}$

Multiply $- 64$ by $- 3$ .

$x = \frac{- 2 \pm \sqrt{4 + 192}}{2 \cdot 16}$

Add $4$ and $192$ .

$x = \frac{- 2 \pm \sqrt{196}}{2 \cdot 16}$

Rewrite $196$ as $14^{2}$ .

$x = \frac{- 2 \pm \sqrt{14^{2}}}{2 \cdot 16}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{- 2 \pm 14}{2 \cdot 16}$

Multiply $2$ by $16$ .

$x = \frac{- 2 \pm 14}{32}$

Simplify $\frac{- 2 \pm 14}{32}$ .

$x = \frac{- 1 \pm 7}{16}$

Step 4

The final answer is the combination of both solutions.

$x = \frac{3}{8}, - \frac{1}{2}$