(1+3t)(t-4)+2(t+4)=4t

Solution for (1+3t)(t-4)+2(t+4)=4t equation:

(1+3t)(t-4)+2(t+4)=4t

We move all terms to the left:

(1+3t)(t-4)+2(t+4)-(4t)=0

We add all the numbers together, and all the variables

(3t+1)(t-4)+2(t+4)-4t=0

We add all the numbers together, and all the variables

-4t+(3t+1)(t-4)+2(t+4)=0

We multiply parentheses

-4t+(3t+1)(t-4)+2t+8=0

We multiply parentheses ..

(+3t^2-12t+t-4)-4t+2t+8=0

We add all the numbers together, and all the variables

(+3t^2-12t+t-4)-2t+8=0

We get rid of parentheses

3t^2-12t+t-2t-4+8=0

We add all the numbers together, and all the variables

3t^2-13t+4=0

a = 3; b = -13; c = +4;
Δ = b2-4ac
Δ = -132-4·3·4
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-11}{2*3}=\frac{2}{6}
=1/3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+11}{2*3}=\frac{24}{6}
=4
$