Solve Using the Quadratic Formula 2x^2=19x+33

$2 x^{2} = 19 x + 33$

Step 1

Move all the expressions to the left side of the equation.

Subtract $19 x$ from both sides of the equation.

$2 x^{2} - 19 x = 33$

Subtract $33$ from both sides of the equation.

$2 x^{2} - 19 x - 33 = 0$

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 $a = 2$ , $b = - 19$ , and $c = - 33$ into the quadratic formula and solve for $x$ .

$\frac{19 \pm \sqrt{{(- 19)}^{2} - 4 \cdot (2 \cdot - 33)}}{2 \cdot 2}$

Step 4

Simplify.

Simplify the numerator.

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

$x = \frac{19 \pm \sqrt{361 - 4 \cdot 2 \cdot - 33}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot - 33$ .

Multiply $- 4$ by $2$ .

$x = \frac{19 \pm \sqrt{361 - 8 \cdot - 33}}{2 \cdot 2}$

Multiply $- 8$ by $- 33$ .

$x = \frac{19 \pm \sqrt{361 + 264}}{2 \cdot 2}$

Add $361$ and $264$ .

$x = \frac{19 \pm \sqrt{625}}{2 \cdot 2}$

Rewrite $625$ as $25^{2}$ .

$x = \frac{19 \pm \sqrt{25^{2}}}{2 \cdot 2}$

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

$x = \frac{19 \pm 25}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{19 \pm 25}{4}$

Step 5

The final answer is the combination of both solutions.

$x = 11, - \frac{3}{2}$