2/3t+12=-2t

Solution for 2/3t+12=-2t equation:

2/3t+12=-2t

We move all terms to the left:

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


Domain of the equation: 3t!=0

t!=0/3

t!=0

t∈R


We get rid of parentheses

2/3t+2t+12=0

We multiply all the terms by the denominator


2t*3t+12*3t+2=0

Wy multiply elements

6t^2+36t+2=0

a = 6; b = 36; c = +2;
Δ = b2-4ac
Δ = 362-4·6·2
Δ = 1248
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:

$\sqrt{\Delta}=\sqrt{1248}=\sqrt{16*78}=\sqrt{16}*\sqrt{78}=4\sqrt{78}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-4\sqrt{78}}{2*6}=\frac{-36-4\sqrt{78}}{12}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+4\sqrt{78}}{2*6}=\frac{-36+4\sqrt{78}}{12}
$