F(t)=t^2+2t-35

Solution for F(t)=t^2+2t-35 equation:

(F)=F^2+2F-35

We move all terms to the left:

(F)-(F^2+2F-35)=0

We get rid of parentheses

-F^2+F-2F+35=0

We add all the numbers together, and all the variables

-1F^2-1F+35=0

a = -1; b = -1; c = +35;
Δ = b2-4ac
Δ = -12-4·(-1)·35
Δ = 141
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{-(-1)-\sqrt{141}}{2*-1}=\frac{1-\sqrt{141}}{-2}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{141}}{2*-1}=\frac{1+\sqrt{141}}{-2}
$