# How do you find the zeros, real and imaginary, of y=-x^2+4x-42 using the quadratic formula?

Jun 7, 2016

The roots are ${x}_{1} = 2 - \sqrt{38} i$ and ${x}_{2} = 2 + \sqrt{38} i$.

#### Explanation:

Function $y = a {x}^{2} + b x + c$ has two roots (zeros) ${x}_{1 , 2} = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$ or one double root if ${b}^{2} = 4 a c$.

In this case $a = - 1 , b = 4 , c = - 42$

So you can blindly plug in those values:

x_{1,2}=(-(4)+-sqrt((4)^2-4(-1)(-42)))/(2(-1))=(-4+-sqrt(16-168))/-2 =(4+-sqrt(-152))/2=(4+-2sqrt(38)i)/2=2+-sqrt(38)i

The roots are ${x}_{1} = 2 - \sqrt{38} i$ and ${x}_{2} = 2 + \sqrt{38} i$.

$\sqrt{38} = 6.164414 \ldots$