K=2x^2+8x

Solution for K=2x^2+8x equation:

=2K^2+8K

We move all terms to the left:

-(2K^2+8K)=0

We get rid of parentheses

-2K^2-8K=0

a = -2; b = -8; c = 0;
Δ = b2-4ac
Δ = -82-4·(-2)·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$K_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$K_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$K_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8}{2*-2}=\frac{0}{-4}
=0
$
$K_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8}{2*-2}=\frac{16}{-4}
=-4
$