F(t)=t^2+6t-18

Solution for F(t)=t^2+6t-18 equation:

(F)=F^2+6F-18

We move all terms to the left:

(F)-(F^2+6F-18)=0

We get rid of parentheses

-F^2+F-6F+18=0

We add all the numbers together, and all the variables

-1F^2-5F+18=0

a = -1; b = -5; c = +18;
Δ = b2-4ac
Δ = -52-4·(-1)·18
Δ = 97
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{97}}{2*-1}=\frac{5-\sqrt{97}}{-2}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{97}}{2*-1}=\frac{5+\sqrt{97}}{-2}
$