18+12t+-3t2=0

Solution for 18+12t+-3t2=0 equation:

18+12t+-3t^2=0

We add all the numbers together, and all the variables

-3t^2+12t=0

a = -3; b = 12; c = 0;
Δ = b2-4ac
Δ = 122-4·(-3)·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12}{2*-3}=\frac{-24}{-6}
=+4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12}{2*-3}=\frac{0}{-6}
=0
$