-4r^2+20r-13=0

Solution for -4r^2+20r-13=0 equation:

-4r^2+20r-13=0

a = -4; b = 20; c = -13;
Δ = b2-4ac
Δ = 202-4·(-4)·(-13)
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-8\sqrt{3}}{2*-4}=\frac{-20-8\sqrt{3}}{-8}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+8\sqrt{3}}{2*-4}=\frac{-20+8\sqrt{3}}{-8}
$