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

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

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

We move all terms to the left:

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


We calculate terms in parentheses: -(3t(t+4)-t), so:

3t(t+4)-t

We add all the numbers together, and all the variables

-1t+3t(t+4)

We multiply parentheses

3t^2-1t+12t

We add all the numbers together, and all the variables

3t^2+11t

Back to the equation:

-(3t^2+11t)


We get rid of parentheses

-3t^2+6t-11t=0

We add all the numbers together, and all the variables

-3t^2-5t=0

a = -3; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·(-3)·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*-3}=\frac{0}{-6}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*-3}=\frac{10}{-6}
=-1+2/3
$