4+1/3t+t=12

Solution for 4+1/3t+t=12 equation:

4+1/3t+t=12

We move all terms to the left:

4+1/3t+t-(12)=0


Domain of the equation: 3t!=0

t!=0/3

t!=0

t∈R


We add all the numbers together, and all the variables

t+1/3t-8=0

We multiply all the terms by the denominator


t*3t-8*3t+1=0

Wy multiply elements

3t^2-24t+1=0

a = 3; b = -24; c = +1;
Δ = b2-4ac
Δ = -242-4·3·1
Δ = 564
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:

$\sqrt{\Delta}=\sqrt{564}=\sqrt{4*141}=\sqrt{4}*\sqrt{141}=2\sqrt{141}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-2\sqrt{141}}{2*3}=\frac{24-2\sqrt{141}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+2\sqrt{141}}{2*3}=\frac{24+2\sqrt{141}}{6}
$