R(x)=x^2+40x

Solution for R(x)=x^2+40x equation:

(R)=R^2+40R

We move all terms to the left:

(R)-(R^2+40R)=0

We get rid of parentheses

-R^2+R-40R=0

We add all the numbers together, and all the variables

-1R^2-39R=0

a = -1; b = -39; c = 0;
Δ = b2-4ac
Δ = -392-4·(-1)·0
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-39)-39}{2*-1}=\frac{0}{-2}
=0
$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-39)+39}{2*-1}=\frac{78}{-2}
=-39
$