2x^2-20x=99

Solution for 2x^2-20x=99 equation:

2x^2-20x=99

We move all terms to the left:

2x^2-20x-(99)=0

a = 2; b = -20; c = -99;
Δ = b2-4ac
Δ = -202-4·2·(-99)
Δ = 1192
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:

$\sqrt{\Delta}=\sqrt{1192}=\sqrt{4*298}=\sqrt{4}*\sqrt{298}=2\sqrt{298}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{298}}{2*2}=\frac{20-2\sqrt{298}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{298}}{2*2}=\frac{20+2\sqrt{298}}{4}
$