5r2+6r-80=0

Solution for 5r2+6r-80=0 equation:

5r^2+6r-80=0

a = 5; b = 6; c = -80;
Δ = b2-4ac
Δ = 62-4·5·(-80)
Δ = 1636
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:

$\sqrt{\Delta}=\sqrt{1636}=\sqrt{4*409}=\sqrt{4}*\sqrt{409}=2\sqrt{409}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{409}}{2*5}=\frac{-6-2\sqrt{409}}{10}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{409}}{2*5}=\frac{-6+2\sqrt{409}}{10}
$