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

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Answer 1 · 2015-07-20T07:34:03

#x=(3+sqrt(19)i)/2, (3-sqrt(19)i)/2#

#x^2-3x=-7#

Get all terms on the left side.

#x^2-3x+7=0# is a quadratic equation in the form #ax^2+bx+c=0#, where #a=1, b=-3, and c=7#.

Quadratic Equation

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

Substitute known values into the formula.

#x=(-(-3)+-sqrt(-3^2-4*1*7))/(2*1)# =

#x=(3+-sqrt(9-28))/2#

#x=(3+-sqrt(-19))/2#

#x=(3+-sqrt(19)i)/2#

Solve for #x#.

#x=(3+sqrt(19)i)/2#

#x=(3-sqrt(19)i)/2#