How do you solve the quadratic using the quadratic formula given #x^2-3x+10=0#?

2 Answers
Nov 9, 2016

#0=x^2-3x+10#

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

#x=[-(-3)+-sqrt((-3)^2-4(1)(10))]/[2(1)]#

Simplify:
#x=[3+-sqrt(-31)]/[2]#

#x=[3+-isqrt31]/2#

Nov 9, 2016

Answer:

#x=(3+sqrt(31)i)/2, (3-sqrt(31)i)/2#

Explanation:

#x^2-3x+10=0# is a quadratic equation in the form #ax^2+bx+c#, where #a=1#, #b=-3#, and #c=10#.

Quadratic Formula

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

#x=(-(-3)+-sqrt((-3)^2-4*1*10))/(2*1)#

#x=(3+-sqrt(9-40))/2#

#x=(3+-sqrt(-31))/2#

#x=(3+sqrt(31)i)/2, (3-sqrt(31)i)/2#