100=(x^2)+48x

Solution for 100=(x^2)+48x equation:

100=(x^2)+48x

We move all terms to the left:

100-((x^2)+48x)=0

We get rid of parentheses

-x^2-48x+100=0

We add all the numbers together, and all the variables

-1x^2-48x+100=0

a = -1; b = -48; c = +100;
Δ = b2-4ac
Δ = -482-4·(-1)·100
Δ = 2704
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2704}=52$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-52}{2*-1}=\frac{-4}{-2}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+52}{2*-1}=\frac{100}{-2}
=-50
$