X^2+(x+4)=20^2

Solution for X^2+(x+4)=20^2 equation:

X^2+(X+4)=20^2

We move all terms to the left:

X^2+(X+4)-(20^2)=0

We add all the numbers together, and all the variables

X^2+(X+4)-400=0

We get rid of parentheses

X^2+X+4-400=0

We add all the numbers together, and all the variables

X^2+X-396=0

a = 1; b = 1; c = -396;
Δ = b2-4ac
Δ = 12-4·1·(-396)
Δ = 1585
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{1585}}{2*1}=\frac{-1-\sqrt{1585}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{1585}}{2*1}=\frac{-1+\sqrt{1585}}{2}
$