How do you find the roots, real and imaginary, of $y= 5x^2-6x +35$ using the quadratic formula?

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Answer 1 · 2017-04-09T15:32:50

$x = 3/5 +- (sqrt(166)i)/5$

Use the quadratic formula $x = (-B +- sqrt(B^2 - 4AC))/(2A)$ when the quadratic equation is in the form $Ax^2 + Bx +C = 0$

For $y = 5x^2 - 6x + 35 = 0, " "A = 5, B = -6, C = 35$

$x = (-B +- sqrt(B^2 - 4AC))/(2A) = (6 +- sqrt(36 - 4(5)(35)))/(2*5)$

$x = (6 +- sqrt(-664))/10 = 3/5 +- sqrt(-4*166)/10 = 3/5 +- (2 sqrt(166) i)/10$

$x = 3/5 +- (sqrt(166)i)/5$

How do you find the roots, real and imaginary, of y= 5x^2-6x +35 y= 5x^2-6x +35 using the quadratic formula?