F(t)=12/3t+4

Solution for F(t)=12/3t+4 equation:

(F)=12/3F+4

We move all terms to the left:

(F)-(12/3F+4)=0


Domain of the equation: 3F+4)!=0

F∈R


We get rid of parentheses

F-12/3F-4=0

We multiply all the terms by the denominator


F*3F-4*3F-12=0

Wy multiply elements

3F^2-12F-12=0

a = 3; b = -12; c = -12;
Δ = b2-4ac
Δ = -122-4·3·(-12)
Δ = 288
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-12\sqrt{2}}{2*3}=\frac{12-12\sqrt{2}}{6}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12\sqrt{2}}{2*3}=\frac{12+12\sqrt{2}}{6}
$