F(x)=(5x-7)(2x-4)

Solution for F(x)=(5x-7)(2x-4) equation:

(F)=(5F-7)(2F-4)

We move all terms to the left:

(F)-((5F-7)(2F-4))=0

We multiply parentheses ..

-((+10F^2-20F-14F+28))+F=0


We calculate terms in parentheses: -((+10F^2-20F-14F+28)), so:

(+10F^2-20F-14F+28)

We get rid of parentheses

10F^2-20F-14F+28

We add all the numbers together, and all the variables

10F^2-34F+28

Back to the equation:

-(10F^2-34F+28)


We add all the numbers together, and all the variables

F-(10F^2-34F+28)=0

We get rid of parentheses

-10F^2+F+34F-28=0

We add all the numbers together, and all the variables

-10F^2+35F-28=0

a = -10; b = 35; c = -28;
Δ = b2-4ac
Δ = 352-4·(-10)·(-28)
Δ = 105
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{-(35)-\sqrt{105}}{2*-10}=\frac{-35-\sqrt{105}}{-20}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+\sqrt{105}}{2*-10}=\frac{-35+\sqrt{105}}{-20}
$