F(x)=(4-x)(7+x)

Solution for F(x)=(4-x)(7+x) equation:

(F)=(4-F)(7+F)

We move all terms to the left:

(F)-((4-F)(7+F))=0

We add all the numbers together, and all the variables

F-((-1F+4)(F+7))=0

We multiply parentheses ..

-((-1F^2-7F+4F+28))+F=0


We calculate terms in parentheses: -((-1F^2-7F+4F+28)), so:

(-1F^2-7F+4F+28)

We get rid of parentheses

-1F^2-7F+4F+28

We add all the numbers together, and all the variables

-1F^2-3F+28

Back to the equation:

-(-1F^2-3F+28)


We get rid of parentheses

1F^2+3F+F-28=0

We add all the numbers together, and all the variables

F^2+4F-28=0

a = 1; b = 4; c = -28;
Δ = b2-4ac
Δ = 42-4·1·(-28)
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8\sqrt{2}}{2*1}=\frac{-4-8\sqrt{2}}{2}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8\sqrt{2}}{2*1}=\frac{-4+8\sqrt{2}}{2}
$