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

Question

Eli Assouline · Accepted Answer

#x# can either equal #3.5 or -1#

#2x^2 - 5x = -7#

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

#2x^2 - 5x + 7= 0#

We can now solve the problem using the quadratic formula.

#ax^2 + bx + c = 0#

#x = (-b+- sqrt(b^2 - 4ac))/(2a)#

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

#x = (5+- sqrt(-5^2 - 4 xx 2 xx 7))/(2 xx 2)#

#x = (5+- sqrt(-25 - 8 xx 7))/4#

#x = (5+- sqrt(-81))/4#

#x = (5+- 9)/4#

#x_1 = (5 + 9)/4#

#x_1 = 14/4#

#color(blue)(x_1 = 3.5#

#x_2 = (5 - 9)/4#

#x_2 = (-4)/4#

#color(blue)(x_2 = -1#

Eva Abramson · Answer

undefined To solve the quadratic equation $2x^2 - 5x = -7$ using the quadratic formula:

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

where $a = 2$, $b = -5$, and $c = -7$.

Substitute the values of $a$, $b$, and $c$ into the formula:

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

$$x = \frac{{5 \pm \sqrt{{25 + 56}}}}{{4}}$$

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

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

Now, we have two solutions:

1. When $x = \frac{{5 + 9}}{{4}}$:

$$x_1 = \frac{{14}}{{4}} = \frac{{7}}{{2}}$$

2. When $x = \frac{{5 - 9}}{{4}}$:

$$x_2 = \frac{{-4}}{{4}} = -1$$

Therefore, the solutions to the equation $2x^2 - 5x = -7$ are $x = \frac{{7}}{{2}}$ and $x = -1$.