5t^2-20t-12=0

Solution for 5t^2-20t-12=0 equation:

5t^2-20t-12=0

a = 5; b = -20; c = -12;
Δ = b2-4ac
Δ = -202-4·5·(-12)
Δ = 640
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:

$\sqrt{\Delta}=\sqrt{640}=\sqrt{64*10}=\sqrt{64}*\sqrt{10}=8\sqrt{10}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-8\sqrt{10}}{2*5}=\frac{20-8\sqrt{10}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+8\sqrt{10}}{2*5}=\frac{20+8\sqrt{10}}{10}
$