How do you find the zeros, real and imaginary, of $y = 75x^2 +100x + 23$ using the quadratic formula?

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Answer 1 · 2017-06-04T16:05:31

$x = -2/3 +- (sqrt(31))/15$

Given: $y = 75x^2 + 100x + 23$

Using the quadratic formula requires the equation to be in the form $Ax^2 + Bx +C = 0$

Quadratic formula: $x = (-B +- sqrt(B^2 - 4AC))/(2A)$

For the given, $A = 75, B = 100, C = 23$ :

$x = (-100 +- sqrt(10,000 - 4*75*23))/(2*75)$

$x = (-100 +- sqrt(10,000 - 6900))/150 = (-100 +- sqrt(3100))/150$

$x = -10/15 +- sqrt(25*4*31)/150 = -10/15 +- (sqrt(25)sqrt(4)sqrt(31))/150$

$x = -2/3 +- (10 sqrt(31))/150 = -2/3 +- (sqrt(31))/15$

How do you find the zeros, real and imaginary, of y = 75x^2 +100x + 23 y = 75x^2 +100x + 23 using the quadratic formula?