400=x^2+12*x

Solution for 400=x^2+12*x equation:

400=x^2+12x

We move all terms to the left:

400-(x^2+12x)=0

We get rid of parentheses

-x^2-12x+400=0

We add all the numbers together, and all the variables

-1x^2-12x+400=0

a = -1; b = -12; c = +400;
Δ = b2-4ac
Δ = -122-4·(-1)·400
Δ = 1744
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:

$\sqrt{\Delta}=\sqrt{1744}=\sqrt{16*109}=\sqrt{16}*\sqrt{109}=4\sqrt{109}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{109}}{2*-1}=\frac{12-4\sqrt{109}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{109}}{2*-1}=\frac{12+4\sqrt{109}}{-2}
$