Solve Using the Quadratic Formula 4x^2-3x-1=0

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

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 $a = 4$ , $b = - 3$ , and $c = - 1$ into the quadratic formula and solve for $x$ .

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

Step 3

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $4$ .

$x = \frac{3 \pm \sqrt{9 - 16 \cdot - 1}}{2 \cdot 4}$

Multiply $- 16$ by $- 1$ .

$x = \frac{3 \pm \sqrt{9 + 16}}{2 \cdot 4}$

Add $9$ and $16$ .

$x = \frac{3 \pm \sqrt{25}}{2 \cdot 4}$

Rewrite $25$ as $5^{2}$ .

$x = \frac{3 \pm \sqrt{5^{2}}}{2 \cdot 4}$

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

$x = \frac{3 \pm 5}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{3 \pm 5}{8}$

Step 4

The final answer is the combination of both solutions.

$x = 1, - \frac{1}{4}$