98=x^2+20x

Solution for 98=x^2+20x equation:

98=x^2+20x

We move all terms to the left:

98-(x^2+20x)=0

We get rid of parentheses

-x^2-20x+98=0

We add all the numbers together, and all the variables

-1x^2-20x+98=0

a = -1; b = -20; c = +98;
Δ = b2-4ac
Δ = -202-4·(-1)·98
Δ = 792
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:

$\sqrt{\Delta}=\sqrt{792}=\sqrt{36*22}=\sqrt{36}*\sqrt{22}=6\sqrt{22}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-6\sqrt{22}}{2*-1}=\frac{20-6\sqrt{22}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+6\sqrt{22}}{2*-1}=\frac{20+6\sqrt{22}}{-2}
$