How do you find the zeros, real and imaginary, of $y=- 5x^2+15x+10$ using the quadratic formula?

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Answer 1 · 2016-02-17T16:49:43

The answers are: $x=(3 pm sqrt(17))/2$ , which are approximately $-0.56$ and $3.56$ .

The quadratic formula says the zeros of $y=ax^2+bx+c$ are $x=(-b pm sqrt(b^2-4ac))/(2a)$ .

For this example, $a=-5$ , $b=15$ , and $c=10$ , so we get:

$x=(-15 pm sqrt( (15)^2-4 * (-5) * 10) )/(2*(-5)) =(15 pm sqrt(225+200))/10$

$=(15 pm sqrt(425))/10=(15 pm sqrt(25)sqrt(17))/10$

$=(15 pm 5sqrt(17))/10=(3 pm sqrt(17))/2$

How do you find the zeros, real and imaginary, of y=- 5x^2+15x+10 y=- 5x^2+15x+10 using the quadratic formula?