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

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

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

We move all terms to the left:

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

We multiply parentheses ..

-((+8F^2+8F-6F-6))+F=0


We calculate terms in parentheses: -((+8F^2+8F-6F-6)), so:

(+8F^2+8F-6F-6)

We get rid of parentheses

8F^2+8F-6F-6

We add all the numbers together, and all the variables

8F^2+2F-6

Back to the equation:

-(8F^2+2F-6)


We add all the numbers together, and all the variables

F-(8F^2+2F-6)=0

We get rid of parentheses

-8F^2+F-2F+6=0

We add all the numbers together, and all the variables

-8F^2-1F+6=0

a = -8; b = -1; c = +6;
Δ = b2-4ac
Δ = -12-4·(-8)·6
Δ = 193
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{193}}{2*-8}=\frac{1-\sqrt{193}}{-16}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{193}}{2*-8}=\frac{1+\sqrt{193}}{-16}
$