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

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Answer 1 · 2017-05-31T07:51:22

#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#

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

#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#