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 = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-14}{2*2}=\frac{-4}{4}
=-1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+14}{2*2}=\frac{24}{4}
=6
$