2r^2-20r-13=0

Solution for 2r^2-20r-13=0 equation:

2r^2-20r-13=0

a = 2; b = -20; c = -13;
Δ = b2-4ac
Δ = -202-4·2·(-13)
Δ = 504
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{504}=\sqrt{36*14}=\sqrt{36}*\sqrt{14}=6\sqrt{14}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-6\sqrt{14}}{2*2}=\frac{20-6\sqrt{14}}{4}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+6\sqrt{14}}{2*2}=\frac{20+6\sqrt{14}}{4}
$