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

Question

1579 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-09T04:37:33

#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#

Answer 2 · 2016-11-09T04:52:40

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

#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#