How do you solve #2x^2 -3x - 7 = 0# using the quadratic formula?

Question

10597 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-25T15:09:35

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

To solve using the quadratic

#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)#