K=5x^2+2x

Solution for K=5x^2+2x equation:

=5K^2+2K

We move all terms to the left:

-(5K^2+2K)=0

We get rid of parentheses

-5K^2-2K=0

a = -5; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·(-5)·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$K_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$K_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$K_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*-5}=\frac{0}{-10}
=0
$
$K_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*-5}=\frac{4}{-10}
=-2/5
$