C(x)=4x^2+2x+K

Solution for C(x)=4x^2+2x+K equation:

(C)=4C^2+2C+

We move all terms to the left:

(C)-(4C^2+2C+)=0

We get rid of parentheses

-4C^2+C-2C-=0

We add all the numbers together, and all the variables

-4C^2-1C=0

a = -4; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·(-4)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$C_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$C_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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