F(2a)=3(2a)^2+7(2a)+5

Solution for F(2a)=3(2a)^2+7(2a)+5 equation:

(2F)=3(2F)^2+7(2F)+5

We move all terms to the left:

(2F)-(3(2F)^2+7(2F)+5)=0

We get rid of parentheses

-32F^2+2F-72F-5=0

We add all the numbers together, and all the variables

-32F^2-70F-5=0

a = -32; b = -70; c = -5;
Δ = b2-4ac
Δ = -702-4·(-32)·(-5)
Δ = 4260
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:

$\sqrt{\Delta}=\sqrt{4260}=\sqrt{4*1065}=\sqrt{4}*\sqrt{1065}=2\sqrt{1065}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-70)-2\sqrt{1065}}{2*-32}=\frac{70-2\sqrt{1065}}{-64}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-70)+2\sqrt{1065}}{2*-32}=\frac{70+2\sqrt{1065}}{-64}
$