How do you solve #x^2 - 4x + 13=0# using the quadratic formula?

1 Answer
Apr 14, 2017

Answer:

Solution: #x= 2+3i , x=2-3i #

Explanation:

Comparing with standard equation #ax^2+bx+c=0#, we get here #a=1, b=-4,c=13 ; D=b^2-4ac= 16-52= -36#.

Discriminant (D) is negative , so roots are complex in nature.

Roots are # x= -b/(2a) +- sqrt (b^2-4ac)/(2a) or x = 4/(2*1) +- sqrt (16-52)/(2*1) = 2+- (6*i)/2 = 2 +-3i#

Solution: #x= 2+3i , x=2-3i # [Ans]