How do you find the zeros, real and imaginary, of y=-4x^2-2x+3 using the quadratic formula?

1 Answer
Rachel
Feb 19, 2016

(-1+--1sqrt(13))/4

Explanation:

The quadratic formula is -b/(2a)+-sqrt(b^2-4*a*c)/(2a)

Let's look at our equation, -4x^2-2x+3, in a more general form: ax^2-bx+c.

So a is -4
b is -2
c is 3.

Now let's plug our variables into the quadratic formula. I like to put parentheses around the variables I'm replacing, just to be extra careful about silly little sign errors.

(-(-2))/(2(-4))+-sqrt((-2)^2-4*(-4)*(3))/(2(-4)).

This becomes 2/-8+-sqrt(4-(-48))/-8, which we can simplify to -1/4+-(2sqrt(13))/-8, or -1/4+-(-1sqrt(13))/4.

The final answer is (-1+--1sqrt(13))/4

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