How do you use the quadratic formula to solve $2x^2-3x+2=0$ ?

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Answer 1 · 2018-03-30T05:22:36

No real solutions. Imaginary solutions: $x=(3+-sqrt(7)i)/4$

The quadratic formula allows us to solve a quadratic equation in the form $ax^2+bx+c$ for $x$ using the following:

$x=(-b+-sqrt(b^4-4ac))/(2a)$ . So, generally, there will be two solutions.

For $2x^2-3x+2, a=2, b=-3, c=2$ , and so

$x=(3+-sqrt((-3)^2-(4)(2)(2)))/(2*2)=(3+-sqrt(9-16))/4=(3+-sqrt(-7))/4$

Due to the negative root, there will be no real solutions, but the imaginary solutions would be

$x=(3+-sqrt(7)i)/4$

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