22500+400x+x^2=400+2x

Solution for 22500+400x+x^2=400+2x equation:

22500+400x+x^2=400+2x

We move all terms to the left:

22500+400x+x^2-(400+2x)=0

We add all the numbers together, and all the variables

x^2+400x-(2x+400)+22500=0

We get rid of parentheses

x^2+400x-2x-400+22500=0

We add all the numbers together, and all the variables

x^2+398x+22100=0

a = 1; b = 398; c = +22100;
Δ = b2-4ac
Δ = 3982-4·1·22100
Δ = 70004
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{70004}=\sqrt{4*17501}=\sqrt{4}*\sqrt{17501}=2\sqrt{17501}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(398)-2\sqrt{17501}}{2*1}=\frac{-398-2\sqrt{17501}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(398)+2\sqrt{17501}}{2*1}=\frac{-398+2\sqrt{17501}}{2}
$