F(x)=4(x+5)(2x-1)

Solution for F(x)=4(x+5)(2x-1) equation:

(F)=4(F+5)(2F-1)

We move all terms to the left:

(F)-(4(F+5)(2F-1))=0

We multiply parentheses ..

-(4(+2F^2-1F+10F-5))+F=0


We calculate terms in parentheses: -(4(+2F^2-1F+10F-5)), so:

4(+2F^2-1F+10F-5)

We multiply parentheses

8F^2-4F+40F-20

We add all the numbers together, and all the variables

8F^2+36F-20

Back to the equation:

-(8F^2+36F-20)


We add all the numbers together, and all the variables

F-(8F^2+36F-20)=0

We get rid of parentheses

-8F^2+F-36F+20=0

We add all the numbers together, and all the variables

-8F^2-35F+20=0

a = -8; b = -35; c = +20;
Δ = b2-4ac
Δ = -352-4·(-8)·20
Δ = 1865
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{1865}}{2*-8}=\frac{35-\sqrt{1865}}{-16}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+\sqrt{1865}}{2*-8}=\frac{35+\sqrt{1865}}{-16}
$