F(t)=2.25(t/5)+9

Solution for F(t)=2.25(t/5)+9 equation:

(F)=2.25(F/5)+9

We move all terms to the left:

(F)-(2.25(F/5)+9)=0

We add all the numbers together, and all the variables

F-(2.25(+F/5)+9)=0

We multiply all the terms by the denominator


F*5)+9)-(2.25(+F=0

We add all the numbers together, and all the variables

F+F*5)+9)-(2.25(=0

Wy multiply elements

5F^2+F=0

a = 5; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·5·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*5}=\frac{-2}{10}
=-1/5
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*5}=\frac{0}{10}
=0
$