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

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

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

We move all terms to the left:

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

We get rid of parentheses

-22F^2+2F+32F-7=0

We add all the numbers together, and all the variables

-22F^2+34F-7=0

a = -22; b = 34; c = -7;
Δ = b2-4ac
Δ = 342-4·(-22)·(-7)
Δ = 540
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:

$\sqrt{\Delta}=\sqrt{540}=\sqrt{36*15}=\sqrt{36}*\sqrt{15}=6\sqrt{15}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-6\sqrt{15}}{2*-22}=\frac{-34-6\sqrt{15}}{-44}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+6\sqrt{15}}{2*-22}=\frac{-34+6\sqrt{15}}{-44}
$