5t^2+10t=120

Solution for 5t^2+10t=120 equation:

5t^2+10t=120

We move all terms to the left:

5t^2+10t-(120)=0

a = 5; b = 10; c = -120;
Δ = b2-4ac
Δ = 102-4·5·(-120)
Δ = 2500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2500}=50$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-50}{2*5}=\frac{-60}{10}
=-6
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+50}{2*5}=\frac{40}{10}
=4
$