-2t^2-10t-12=0

Solution for -2t^2-10t-12=0 equation:

-2t^2-10t-12=0

a = -2; b = -10; c = -12;
Δ = b2-4ac
Δ = -102-4·(-2)·(-12)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2}{2*-2}=\frac{8}{-4}
=-2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2}{2*-2}=\frac{12}{-4}
=-3
$