-5t^2+12t+2=0

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

-5t^2+12t+2=0

a = -5; b = 12; c = +2;
Δ = b2-4ac
Δ = 122-4·(-5)·2
Δ = 184
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:

$\sqrt{\Delta}=\sqrt{184}=\sqrt{4*46}=\sqrt{4}*\sqrt{46}=2\sqrt{46}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{46}}{2*-5}=\frac{-12-2\sqrt{46}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{46}}{2*-5}=\frac{-12+2\sqrt{46}}{-10}
$