2t^2+t-120=0

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

2t^2+t-120=0

a = 2; b = 1; c = -120;
Δ = b2-4ac
Δ = 12-4·2·(-120)
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-31}{2*2}=\frac{-32}{4}
=-8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+31}{2*2}=\frac{30}{4}
=7+1/2
$