0.2t^2+10t-960=0

Solution for 0.2t^2+10t-960=0 equation:

0.2t^2+10t-960=0

a = 0.2; b = 10; c = -960;
Δ = b2-4ac
Δ = 102-4·0.2·(-960)
Δ = 868
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:

$\sqrt{\Delta}=\sqrt{868}=\sqrt{4*217}=\sqrt{4}*\sqrt{217}=2\sqrt{217}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{217}}{2*0.2}=\frac{-10-2\sqrt{217}}{0.4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{217}}{2*0.2}=\frac{-10+2\sqrt{217}}{0.4}
$