F(t)=5t^2+30t+10

Solution for F(t)=5t^2+30t+10 equation:

(F)=5F^2+30F+10

We move all terms to the left:

(F)-(5F^2+30F+10)=0

We get rid of parentheses

-5F^2+F-30F-10=0

We add all the numbers together, and all the variables

-5F^2-29F-10=0

a = -5; b = -29; c = -10;
Δ = b2-4ac
Δ = -292-4·(-5)·(-10)
Δ = 641
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{-(-29)-\sqrt{641}}{2*-5}=\frac{29-\sqrt{641}}{-10}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+\sqrt{641}}{2*-5}=\frac{29+\sqrt{641}}{-10}
$