130=21x+x^2

Solution for 130=21x+x^2 equation:

130=21x+x^2

We move all terms to the left:

130-(21x+x^2)=0

We get rid of parentheses

-x^2-21x+130=0

We add all the numbers together, and all the variables

-1x^2-21x+130=0

a = -1; b = -21; c = +130;
Δ = b2-4ac
Δ = -212-4·(-1)·130
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-31}{2*-1}=\frac{-10}{-2}
=+5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+31}{2*-1}=\frac{52}{-2}
=-26
$