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

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

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

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

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

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