F(x)=3x^2-(4x+8)

Solution for F(x)=3x^2-(4x+8) equation:

(F)=3F^2-(4F+8)

We move all terms to the left:

(F)-(3F^2-(4F+8))=0


We calculate terms in parentheses: -(3F^2-(4F+8)), so:

3F^2-(4F+8)

We get rid of parentheses

3F^2-4F-8

Back to the equation:

-(3F^2-4F-8)


We get rid of parentheses

-3F^2+F+4F+8=0

We add all the numbers together, and all the variables

-3F^2+5F+8=0

a = -3; b = 5; c = +8;
Δ = b2-4ac
Δ = 52-4·(-3)·8
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-11}{2*-3}=\frac{-16}{-6}
=2+2/3
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+11}{2*-3}=\frac{6}{-6}
=-1
$