Solve Using the Quadratic Formula 3x^2-14x+8=0

$3 x^{2} - 14 x + 8 = 0$

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 $a = 3$ , $b = - 14$ , and $c = 8$ into the quadratic formula and solve for $x$ .

$\frac{14 \pm \sqrt{{(- 14)}^{2} - 4 \cdot (3 \cdot 8)}}{2 \cdot 3}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 14$ to the power of $2$ .

$x = \frac{14 \pm \sqrt{196 - 4 \cdot 3 \cdot 8}}{2 \cdot 3}$

Multiply $- 4 \cdot 3 \cdot 8$ .

Multiply $- 4$ by $3$ .

$x = \frac{14 \pm \sqrt{196 - 12 \cdot 8}}{2 \cdot 3}$

Multiply $- 12$ by $8$ .

$x = \frac{14 \pm \sqrt{196 - 96}}{2 \cdot 3}$

Subtract $96$ from $196$ .

$x = \frac{14 \pm \sqrt{100}}{2 \cdot 3}$

Rewrite $100$ as $10^{2}$ .

$x = \frac{14 \pm \sqrt{10^{2}}}{2 \cdot 3}$

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

$x = \frac{14 \pm 10}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{14 \pm 10}{6}$

Simplify $\frac{14 \pm 10}{6}$ .

$x = \frac{7 \pm 5}{3}$

Step 4

The final answer is the combination of both solutions.

$x = 4, \frac{2}{3}$