Solving quadratics with imaginary solutions

Answer:
x = (1+2isqrt6)/5, (1-2isqrt6)/5
Step-by-step explanation:
10x^2-4x+10=0
a=10, b=-4, c=10
by quadratic formula: (you forgot to negate b)
x = (- (-4) +- sqrt(16-4*10*10))/(2*10)
= (4+-sqrt(-384))/20
Here we are simplifying the radical:
= (4+- (i)sqrt(64*6))/20
That was because sqrt(-n) is the same as (i * sqrt(n))
= (4+- (i)sqrt(8^2*6))/20
= (4+- (8i)sqrt(6))/20
= (1+-2isqrt6)/5
Two solutions to +- expression:
x = (1 + 2isqrt6)/5, (1-2isqrt6)/5
Hope this helps