3(t+18)=-3(2-t)t

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

3(t+18)=-3(2-t)t

We move all terms to the left:

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

We add all the numbers together, and all the variables

3(t+18)-(-3(-1t+2)t)=0

We multiply parentheses

3t-(-3(-1t+2)t)+54=0


We calculate terms in parentheses: -(-3(-1t+2)t), so:

-3(-1t+2)t

We multiply parentheses

3t^2-6t

Back to the equation:

-(3t^2-6t)


We get rid of parentheses

-3t^2+3t+6t+54=0

We add all the numbers together, and all the variables

-3t^2+9t+54=0

a = -3; b = 9; c = +54;
Δ = b2-4ac
Δ = 92-4·(-3)·54
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-27}{2*-3}=\frac{-36}{-6}
=+6
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+27}{2*-3}=\frac{18}{-6}
=-3
$