15t^2+120t+1=0

Solution for 15t^2+120t+1=0 equation:

15t^2+120t+1=0

a = 15; b = 120; c = +1;
Δ = b2-4ac
Δ = 1202-4·15·1
Δ = 14340
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:

$\sqrt{\Delta}=\sqrt{14340}=\sqrt{4*3585}=\sqrt{4}*\sqrt{3585}=2\sqrt{3585}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(120)-2\sqrt{3585}}{2*15}=\frac{-120-2\sqrt{3585}}{30}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(120)+2\sqrt{3585}}{2*15}=\frac{-120+2\sqrt{3585}}{30}
$