4t^2+16t=120

Solution for 4t^2+16t=120 equation:

4t^2+16t=120

We move all terms to the left:

4t^2+16t-(120)=0

a = 4; b = 16; c = -120;
Δ = b2-4ac
Δ = 162-4·4·(-120)
Δ = 2176
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:

$\sqrt{\Delta}=\sqrt{2176}=\sqrt{64*34}=\sqrt{64}*\sqrt{34}=8\sqrt{34}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8\sqrt{34}}{2*4}=\frac{-16-8\sqrt{34}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8\sqrt{34}}{2*4}=\frac{-16+8\sqrt{34}}{8}
$