3t^2+20t+12t=0

Solution for 3t^2+20t+12t=0 equation:

3t^2+20t+12t=0

We add all the numbers together, and all the variables

3t^2+32t=0

a = 3; b = 32; c = 0;
Δ = b2-4ac
Δ = 322-4·3·0
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-32}{2*3}=\frac{-64}{6}
=-10+2/3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+32}{2*3}=\frac{0}{6}
=0
$