x^2+98=20x

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

x^2+98=20x

We move all terms to the left:

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

a = 1; b = -20; c = +98;
Δ = b2-4ac
Δ = -202-4·1·98
Δ = 8
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:

$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{2}}{2*1}=\frac{20-2\sqrt{2}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{2}}{2*1}=\frac{20+2\sqrt{2}}{2}
$