3(t-4)+6=3t(2t-6)

Solution for 3(t-4)+6=3t(2t-6) equation:

3(t-4)+6=3t(2t-6)

We move all terms to the left:

3(t-4)+6-(3t(2t-6))=0

We multiply parentheses

3t-(3t(2t-6))-12+6=0


We calculate terms in parentheses: -(3t(2t-6)), so:

3t(2t-6)

We multiply parentheses

6t^2-18t

Back to the equation:

-(6t^2-18t)


We add all the numbers together, and all the variables

3t-(6t^2-18t)-6=0

We get rid of parentheses

-6t^2+3t+18t-6=0

We add all the numbers together, and all the variables

-6t^2+21t-6=0

a = -6; b = 21; c = -6;
Δ = b2-4ac
Δ = 212-4·(-6)·(-6)
Δ = 297
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{297}=\sqrt{9*33}=\sqrt{9}*\sqrt{33}=3\sqrt{33}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-3\sqrt{33}}{2*-6}=\frac{-21-3\sqrt{33}}{-12}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+3\sqrt{33}}{2*-6}=\frac{-21+3\sqrt{33}}{-12}
$