-5t^2+20t+6=0

Solution for -5t^2+20t+6=0 equation:

-5t^2+20t+6=0

a = -5; b = 20; c = +6;
Δ = b2-4ac
Δ = 202-4·(-5)·6
Δ = 520
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{520}=\sqrt{4*130}=\sqrt{4}*\sqrt{130}=2\sqrt{130}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{130}}{2*-5}=\frac{-20-2\sqrt{130}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{130}}{2*-5}=\frac{-20+2\sqrt{130}}{-10}
$