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

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

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

We move all terms to the left:

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

We get rid of parentheses

-42F^2+2F-32F=0

We add all the numbers together, and all the variables

-42F^2-30F=0

a = -42; b = -30; c = 0;
Δ = b2-4ac
Δ = -302-4·(-42)·0
Δ = 900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{900}=30$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-30}{2*-42}=\frac{0}{-84}
=0
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+30}{2*-42}=\frac{60}{-84}
=-5/7
$