F(x)=(2x+1)(x+5)

Solution for F(x)=(2x+1)(x+5) equation:

(F)=(2F+1)(F+5)

We move all terms to the left:

(F)-((2F+1)(F+5))=0

We multiply parentheses ..

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


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

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

We get rid of parentheses

2F^2+10F+F+5

We add all the numbers together, and all the variables

2F^2+11F+5

Back to the equation:

-(2F^2+11F+5)


We add all the numbers together, and all the variables

F-(2F^2+11F+5)=0

We get rid of parentheses

-2F^2+F-11F-5=0

We add all the numbers together, and all the variables

-2F^2-10F-5=0

a = -2; b = -10; c = -5;
Δ = b2-4ac
Δ = -102-4·(-2)·(-5)
Δ = 60
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:

$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{15}}{2*-2}=\frac{10-2\sqrt{15}}{-4}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{15}}{2*-2}=\frac{10+2\sqrt{15}}{-4}
$