How do you find the zeros, real and imaginary, of $y= 8x^2+3x-24$ using the quadratic formula?

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Answer 1 · 2015-11-21T05:21:25

$x=(-3+-sqrt777)/16$

The quadratic formula states that for a quadratic equation $ax^2+bx+c$ , $color(red)(x=(-b+-sqrt(b^2-4ac))/(2a)$ .

We know from your equation that: $a=8,b=3,c=-24$

$x=(-3+-sqrt(3^2-4(8)(-24)))/(2(8))=(-3+-sqrt777)/16$

$777$ is prime. These are your fully simplified answers, which are irrational but real.

How do you find the zeros, real and imaginary, of y= 8x^2+3x-24y= 8x^2+3x-24 using the quadratic formula?