Solve Using the Quadratic Formula 10x^2-19x+6=0

$10 x^{2} - 19 x + 6 = 0$

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 $a = 10$ , $b = - 19$ , and $c = 6$ into the quadratic formula and solve for $x$ .

$\frac{19 \pm \sqrt{{(- 19)}^{2} - 4 \cdot (10 \cdot 6)}}{2 \cdot 10}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 19$ to the power of $2$ .

$x = \frac{19 \pm \sqrt{361 - 4 \cdot 10 \cdot 6}}{2 \cdot 10}$

Multiply $- 4 \cdot 10 \cdot 6$ .

Multiply $- 4$ by $10$ .

$x = \frac{19 \pm \sqrt{361 - 40 \cdot 6}}{2 \cdot 10}$

Multiply $- 40$ by $6$ .

$x = \frac{19 \pm \sqrt{361 - 240}}{2 \cdot 10}$

Subtract $240$ from $361$ .

$x = \frac{19 \pm \sqrt{121}}{2 \cdot 10}$

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

$x = \frac{19 \pm \sqrt{11^{2}}}{2 \cdot 10}$

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

$x = \frac{19 \pm 11}{2 \cdot 10}$

Multiply $2$ by $10$ .

$x = \frac{19 \pm 11}{20}$

Step 4

The final answer is the combination of both solutions.

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