400=42(x)+x^2

Solution for 400=42(x)+x^2 equation:

400=42(x)+x^2

We move all terms to the left:

400-(42(x)+x^2)=0

We get rid of parentheses

-x^2-42x+400=0

We add all the numbers together, and all the variables

-1x^2-42x+400=0

a = -1; b = -42; c = +400;
Δ = b2-4ac
Δ = -422-4·(-1)·400
Δ = 3364
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3364}=58$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-58}{2*-1}=\frac{-16}{-2}
=+8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+58}{2*-1}=\frac{100}{-2}
=-50
$