Solve Using the Quadratic Formula 2x^2-3x=5

$2 x^{2} - 3 x = 5$

Step 1

Subtract $5$ from both sides of the equation.

$2 x^{2} - 3 x - 5 = 0$

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 $a = 2$ , $b = - 3$ , and $c = - 5$ into the quadratic formula and solve for $x$ .

$\frac{3 \pm \sqrt{{(- 3)}^{2} - 4 \cdot (2 \cdot - 5)}}{2 \cdot 2}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 3$ to the power of $2$ .

$x = \frac{3 \pm \sqrt{9 - 4 \cdot 2 \cdot - 5}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot - 5$ .

Multiply $- 4$ by $2$ .

$x = \frac{3 \pm \sqrt{9 - 8 \cdot - 5}}{2 \cdot 2}$

Multiply $- 8$ by $- 5$ .

$x = \frac{3 \pm \sqrt{9 + 40}}{2 \cdot 2}$

Add $9$ and $40$ .

$x = \frac{3 \pm \sqrt{49}}{2 \cdot 2}$

Rewrite $49$ as $7^{2}$ .

$x = \frac{3 \pm \sqrt{7^{2}}}{2 \cdot 2}$

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

$x = \frac{3 \pm 7}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{3 \pm 7}{4}$

Step 5

The final answer is the combination of both solutions.

$x = \frac{5}{2}, - 1$