F(x)=(x+5)(x+2)

Solution for F(x)=(x+5)(x+2) equation:

(F)=(F+5)(F+2)

We move all terms to the left:

(F)-((F+5)(F+2))=0

We multiply parentheses ..

-((+F^2+2F+5F+10))+F=0


We calculate terms in parentheses: -((+F^2+2F+5F+10)), so:

(+F^2+2F+5F+10)

We get rid of parentheses

F^2+2F+5F+10

We add all the numbers together, and all the variables

F^2+7F+10

Back to the equation:

-(F^2+7F+10)


We add all the numbers together, and all the variables

F-(F^2+7F+10)=0

We get rid of parentheses

-F^2+F-7F-10=0

We add all the numbers together, and all the variables

-1F^2-6F-10=0

a = -1; b = -6; c = -10;
Δ = b2-4ac
Δ = -62-4·(-1)·(-10)
Δ = -4
Delta is less than zero, so there is no solution for the equation