Solve Using the Quadratic Formula 2x^2+35=17x

$2 x^{2} + 35 = 17 x$

Step 1

Subtract $17 x$ from both sides of the equation.

$2 x^{2} + 35 - 17 x = 0$

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = 2$ , $b = - 17$ , and $c = 35$ into the quadratic formula and solve for $x$ .

$\frac{17 \pm \sqrt{{(- 17)}^{2} - 4 \cdot (2 \cdot 35)}}{2 \cdot 2}$

Step 4

Simplify.

Simplify the numerator.

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

$x = \frac{17 \pm \sqrt{289 - 4 \cdot 2 \cdot 35}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot 35$ .

Multiply $- 4$ by $2$ .

$x = \frac{17 \pm \sqrt{289 - 8 \cdot 35}}{2 \cdot 2}$

Multiply $- 8$ by $35$ .

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

Subtract $280$ from $289$ .

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

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

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

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

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

Multiply $2$ by $2$ .

$x = \frac{17 \pm 3}{4}$

Step 5

The final answer is the combination of both solutions.

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