F(4y)=2(4y)2-3(4y)+12

Solution for F(4y)=2(4y)2-3(4y)+12 equation:

(4F)=2(4F)2-3(4F)+12

We move all terms to the left:

(4F)-(2(4F)2-3(4F)+12)=0

We add all the numbers together, and all the variables

-(+24F^2-34F+12)+4F=0

We get rid of parentheses

-24F^2+34F+4F-12=0

We add all the numbers together, and all the variables

-24F^2+38F-12=0

a = -24; b = 38; c = -12;
Δ = b2-4ac
Δ = 382-4·(-24)·(-12)
Δ = 292
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:

$\sqrt{\Delta}=\sqrt{292}=\sqrt{4*73}=\sqrt{4}*\sqrt{73}=2\sqrt{73}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(38)-2\sqrt{73}}{2*-24}=\frac{-38-2\sqrt{73}}{-48}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(38)+2\sqrt{73}}{2*-24}=\frac{-38+2\sqrt{73}}{-48}
$