Solve Using the Quadratic Formula 5x^2-13x+6=0

$5 x^{2} - 13 x + 6 = 0$

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 $a = 5$ , $b = - 13$ , and $c = 6$ into the quadratic formula and solve for $x$ .

$\frac{13 \pm \sqrt{{(- 13)}^{2} - 4 \cdot (5 \cdot 6)}}{2 \cdot 5}$

Step 3

Simplify.

Simplify the numerator.

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

$x = \frac{13 \pm \sqrt{169 - 4 \cdot 5 \cdot 6}}{2 \cdot 5}$

Multiply $- 4 \cdot 5 \cdot 6$ .

Multiply $- 4$ by $5$ .

$x = \frac{13 \pm \sqrt{169 - 20 \cdot 6}}{2 \cdot 5}$

Multiply $- 20$ by $6$ .

$x = \frac{13 \pm \sqrt{169 - 120}}{2 \cdot 5}$

Subtract $120$ from $169$ .

$x = \frac{13 \pm \sqrt{49}}{2 \cdot 5}$

Rewrite $49$ as $7^{2}$ .

$x = \frac{13 \pm \sqrt{7^{2}}}{2 \cdot 5}$

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

$x = \frac{13 \pm 7}{2 \cdot 5}$

Multiply $2$ by $5$ .

$x = \frac{13 \pm 7}{10}$

Step 4

The final answer is the combination of both solutions.

$x = 2, \frac{3}{5}$