How do you find the roots, real and imaginary, of $y= 2x^2 - 3x + 2$ using the quadratic formula?

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Answer 1 · 2016-04-05T10:06:27

$x_1=(3+sqrt7 i)/4$

$x_2=(3-sqrt7 i)/4$

$x=(-b+-sqrt(b^2-4ac))/(2a)$

we have to identify correctly our real number coefficients a, b ,c

set y=0

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

let $a=2$ and $b=-3$ and $c=2$

$x=(-b+-sqrt(b^2-4ac))/(2a)$

$x=(--3+-sqrt((-3)^2-4(2)(2)))/(2(2))$

$x_1=(3+sqrt7 i)/4$

$x_2=(3-sqrt7 i)/4$

God bless....I hope the explanation is useful.

How do you find the roots, real and imaginary, of y= 2x^2 - 3x + 2 y= 2x^2 - 3x + 2 using the quadratic formula?