F(t)=1/2t-3

Solution for F(t)=1/2t-3 equation:

(F)=1/2F-3

We move all terms to the left:

(F)-(1/2F-3)=0


Domain of the equation: 2F-3)!=0

F∈R


We get rid of parentheses

F-1/2F+3=0

We multiply all the terms by the denominator


F*2F+3*2F-1=0

Wy multiply elements

2F^2+6F-1=0

a = 2; b = 6; c = -1;
Δ = b2-4ac
Δ = 62-4·2·(-1)
Δ = 44
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:

$\sqrt{\Delta}=\sqrt{44}=\sqrt{4*11}=\sqrt{4}*\sqrt{11}=2\sqrt{11}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{11}}{2*2}=\frac{-6-2\sqrt{11}}{4}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{11}}{2*2}=\frac{-6+2\sqrt{11}}{4}
$