3k^2+80k-80=0

Solution for 3k^2+80k-80=0 equation:

3k^2+80k-80=0

a = 3; b = 80; c = -80;
Δ = b2-4ac
Δ = 802-4·3·(-80)
Δ = 7360
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{7360}=\sqrt{64*115}=\sqrt{64}*\sqrt{115}=8\sqrt{115}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-8\sqrt{115}}{2*3}=\frac{-80-8\sqrt{115}}{6}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+8\sqrt{115}}{2*3}=\frac{-80+8\sqrt{115}}{6}
$