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

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Answer 1 · 2017-03-18T13:41:02

The solutions are #S={3/4+sqrt15/4i,3/4-sqrt15/4i}#

The quadratic equation is

#ax^2+bx+c=0#

Here, we have

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

The discriminant is

#Delta=b^2-4ac=9-4*2*3=9-24=-15#

As #Delta<0#, there are no real roots.

The roots are imaginary

#x=(-b+-sqrtDelta)/(2a)#

#x=(3+-isqrt15)/(4)#

#x_1=3/4+sqrt15/4i#

#x_2=3/4-sqrt15/4i#