2r^2+3r-65=0

Solution for 2r^2+3r-65=0 equation:

2r^2+3r-65=0

a = 2; b = 3; c = -65;
Δ = b2-4ac
Δ = 32-4·2·(-65)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-23}{2*2}=\frac{-26}{4}
=-6+1/2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+23}{2*2}=\frac{20}{4}
=5
$