20r^2+39r-6=0

Solution for 20r^2+39r-6=0 equation:

20r^2+39r-6=0

a = 20; b = 39; c = -6;
Δ = b2-4ac
Δ = 392-4·20·(-6)
Δ = 2001
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(39)-\sqrt{2001}}{2*20}=\frac{-39-\sqrt{2001}}{40}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+\sqrt{2001}}{2*20}=\frac{-39+\sqrt{2001}}{40}
$