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.