F(t)=6t2+19t-7

Solution for F(t)=6t2+19t-7 equation:

(F)=6F^2+19F-7

We move all terms to the left:

(F)-(6F^2+19F-7)=0

We get rid of parentheses

-6F^2+F-19F+7=0

We add all the numbers together, and all the variables

-6F^2-18F+7=0

a = -6; b = -18; c = +7;
Δ = b2-4ac
Δ = -182-4·(-6)·7
Δ = 492
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:

$\sqrt{\Delta}=\sqrt{492}=\sqrt{4*123}=\sqrt{4}*\sqrt{123}=2\sqrt{123}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{123}}{2*-6}=\frac{18-2\sqrt{123}}{-12}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{123}}{2*-6}=\frac{18+2\sqrt{123}}{-12}
$