2t^2+30t-120=0

Solution for 2t^2+30t-120=0 equation:

2t^2+30t-120=0

a = 2; b = 30; c = -120;
Δ = b2-4ac
Δ = 302-4·2·(-120)
Δ = 1860
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:

$\sqrt{\Delta}=\sqrt{1860}=\sqrt{4*465}=\sqrt{4}*\sqrt{465}=2\sqrt{465}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-2\sqrt{465}}{2*2}=\frac{-30-2\sqrt{465}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+2\sqrt{465}}{2*2}=\frac{-30+2\sqrt{465}}{4}
$