Solve Using the Quadratic Formula 4x^2-13x+3=0

$4 x^{2} - 13 x + 3 = 0$

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 $a = 4$ , $b = - 13$ , and $c = 3$ into the quadratic formula and solve for $x$ .

$\frac{13 \pm \sqrt{{(- 13)}^{2} - 4 \cdot (4 \cdot 3)}}{2 \cdot 4}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 13$ to the power of $2$ .

$x = \frac{13 \pm \sqrt{169 - 4 \cdot 4 \cdot 3}}{2 \cdot 4}$

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

Multiply $- 4$ by $4$ .

$x = \frac{13 \pm \sqrt{169 - 16 \cdot 3}}{2 \cdot 4}$

Multiply $- 16$ by $3$ .

$x = \frac{13 \pm \sqrt{169 - 48}}{2 \cdot 4}$

Subtract $48$ from $169$ .

$x = \frac{13 \pm \sqrt{121}}{2 \cdot 4}$

Rewrite $121$ as $11^{2}$ .

$x = \frac{13 \pm \sqrt{11^{2}}}{2 \cdot 4}$

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

$x = \frac{13 \pm 11}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{13 \pm 11}{8}$

Step 4

The final answer is the combination of both solutions.

$x = 3, \frac{1}{4}$