1.2t^2-4t-8=0

Solution for 1.2t^2-4t-8=0 equation:

1.2t^2-4t-8=0

a = 1.2; b = -4; c = -8;
Δ = b2-4ac
Δ = -42-4·1.2·(-8)
Δ = 54.4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-\sqrt{54.4}}{2*1.2}=\frac{4-\sqrt{54.4}}{2.4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+\sqrt{54.4}}{2*1.2}=\frac{4+\sqrt{54.4}}{2.4}
$