# How do you solve 2x^2-5x=-7 using the quadratic formula?

May 31, 2017

$x$ can either equal $3.5 \mathmr{and} - 1$

#### Explanation:

$2 {x}^{2} - 5 x = - 7$

To make the answer $0$, which we need in a quadratic equation, we can just move the $- 7$ to the opposite side.

$2 {x}^{2} - 5 x + 7 = 0$

Now we can use the Quadratic formula to solve the question.

$a {x}^{2} + b x + c = 0$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$a = 2$
$b = - 5$
$c = 7$

$x = \frac{5 \pm \sqrt{- {5}^{2} - 4 \times 2 \times 7}}{2 \times 2}$

$x = \frac{5 \pm \sqrt{- 25 - 8 \times 7}}{4}$

$x = \frac{5 \pm \sqrt{- 81}}{4}$

$x = \frac{5 \pm 9}{4}$

${x}_{1} = \frac{5 + 9}{4}$

${x}_{1} = \frac{14}{4}$

color(blue)(x_1 = 3.5

${x}_{2} = \frac{5 - 9}{4}$

${x}_{2} = \frac{- 4}{4}$

color(blue)(x_2 = -1