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

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

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

We move all terms to the left:

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

We multiply parentheses ..

-(5(+F^2-4F+5F-20)2)+F=0


We calculate terms in parentheses: -(5(+F^2-4F+5F-20)2), so:

5(+F^2-4F+5F-20)2

We multiply parentheses

10F^2-40F+50F-200

We add all the numbers together, and all the variables

10F^2+10F-200

Back to the equation:

-(10F^2+10F-200)


We add all the numbers together, and all the variables

F-(10F^2+10F-200)=0

We get rid of parentheses

-10F^2+F-10F+200=0

We add all the numbers together, and all the variables

-10F^2-9F+200=0

a = -10; b = -9; c = +200;
Δ = b2-4ac
Δ = -92-4·(-10)·200
Δ = 8081
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{-(-9)-\sqrt{8081}}{2*-10}=\frac{9-\sqrt{8081}}{-20}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{8081}}{2*-10}=\frac{9+\sqrt{8081}}{-20}
$