16t^2+360t=0

Solution for 16t^2+360t=0 equation:

16t^2+360t=0

a = 16; b = 360; c = 0;
Δ = b2-4ac
Δ = 3602-4·16·0
Δ = 129600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{129600}=360$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(360)-360}{2*16}=\frac{-720}{32}
=-22+1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(360)+360}{2*16}=\frac{0}{32}
=0
$