How do you solve #2x^2 - 3x = 7# using the quadratic formula?

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Answer 1 · 2016-03-22T15:54:26

#x = -1/2+sqrt(5)# or #x = -1/2-sqrt(5)#

Begin by setting the equation equal to zero.

#2x^2-3x -7=0#

Identify the a, b, and c terms

#ax^2-bx -c=0#
#a = 2#
#b = -3#
#c = -7#

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

#x = (-2(+/-)sqrt(2^2-4(-3)(-7))/(2(2)))#

#x = (-2(+/-)sqrt(4-84))/(4)#

#x = (-2(+/-)sqrt(-80))/(4)# simplify the radical #i = sqrt(-1)#

#x = (-2(+/-)4isqrt(5))/(4)# simplify the fractions

#x = -1/2+sqrt(5)# or #x = -1/2-sqrt(5)#