F(5a)=3(5a)^2+4(5a)-2

Solution for F(5a)=3(5a)^2+4(5a)-2 equation:

(5F)=3(5F)^2+4(5F)-2

We move all terms to the left:

(5F)-(3(5F)^2+4(5F)-2)=0

We get rid of parentheses

-35F^2+5F-45F+2=0

We add all the numbers together, and all the variables

-35F^2-40F+2=0

a = -35; b = -40; c = +2;
Δ = b2-4ac
Δ = -402-4·(-35)·2
Δ = 1880
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:

$\sqrt{\Delta}=\sqrt{1880}=\sqrt{4*470}=\sqrt{4}*\sqrt{470}=2\sqrt{470}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-2\sqrt{470}}{2*-35}=\frac{40-2\sqrt{470}}{-70}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+2\sqrt{470}}{2*-35}=\frac{40+2\sqrt{470}}{-70}
$