F(x)=6x^2+4x-5

Solution for F(x)=6x^2+4x-5 equation:

(F)=6F^2+4F-5

We move all terms to the left:

(F)-(6F^2+4F-5)=0

We get rid of parentheses

-6F^2+F-4F+5=0

We add all the numbers together, and all the variables

-6F^2-3F+5=0

a = -6; b = -3; c = +5;
Δ = b2-4ac
Δ = -32-4·(-6)·5
Δ = 129
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{-(-3)-\sqrt{129}}{2*-6}=\frac{3-\sqrt{129}}{-12}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{129}}{2*-6}=\frac{3+\sqrt{129}}{-12}
$