Solve Using the Quadratic Formula 5x^2-17x+14=0

$5 x^{2} - 17 x + 14 = 0$

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 $a = 5$ , $b = - 17$ , and $c = 14$ into the quadratic formula and solve for $x$ .

$\frac{17 \pm \sqrt{{(- 17)}^{2} - 4 \cdot (5 \cdot 14)}}{2 \cdot 5}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 17$ to the power of $2$ .

$x = \frac{17 \pm \sqrt{289 - 4 \cdot 5 \cdot 14}}{2 \cdot 5}$

Multiply $- 4 \cdot 5 \cdot 14$ .

Multiply $- 4$ by $5$ .

$x = \frac{17 \pm \sqrt{289 - 20 \cdot 14}}{2 \cdot 5}$

Multiply $- 20$ by $14$ .

$x = \frac{17 \pm \sqrt{289 - 280}}{2 \cdot 5}$

Subtract $280$ from $289$ .

$x = \frac{17 \pm \sqrt{9}}{2 \cdot 5}$

Rewrite $9$ as $3^{2}$ .

$x = \frac{17 \pm \sqrt{3^{2}}}{2 \cdot 5}$

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

$x = \frac{17 \pm 3}{2 \cdot 5}$

Multiply $2$ by $5$ .

$x = \frac{17 \pm 3}{10}$

Step 4

The final answer is the combination of both solutions.

$x = 2, \frac{7}{5}$