18t-(2t(3t+12))+72=0

Solution for 18t-(2t(3t+12))+72=0 equation:

18t-(2t(3t+12))+72=0


We calculate terms in parentheses: -(2t(3t+12)), so:

2t(3t+12)

We multiply parentheses

6t^2+24t

Back to the equation:

-(6t^2+24t)


We get rid of parentheses

-6t^2+18t-24t+72=0

We add all the numbers together, and all the variables

-6t^2-6t+72=0

a = -6; b = -6; c = +72;
Δ = b2-4ac
Δ = -62-4·(-6)·72
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-42}{2*-6}=\frac{-36}{-12}
=+3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+42}{2*-6}=\frac{48}{-12}
=-4
$