2r^2-15r+20=0

Solution for 2r^2-15r+20=0 equation:

2r^2-15r+20=0

a = 2; b = -15; c = +20;
Δ = b2-4ac
Δ = -152-4·2·20
Δ = 65
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{-(-15)-\sqrt{65}}{2*2}=\frac{15-\sqrt{65}}{4}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{65}}{2*2}=\frac{15+\sqrt{65}}{4}
$