F(x)=-3(x+1)(x-5)

Solution for F(x)=-3(x+1)(x-5) equation:

(F)=-3(F+1)(F-5)

We move all terms to the left:

(F)-(-3(F+1)(F-5))=0

We multiply parentheses ..

-(-3(+F^2-5F+F-5))+F=0


We calculate terms in parentheses: -(-3(+F^2-5F+F-5)), so:

-3(+F^2-5F+F-5)

We multiply parentheses

-3F^2+15F-3F+15

We add all the numbers together, and all the variables

-3F^2+12F+15

Back to the equation:

-(-3F^2+12F+15)


We get rid of parentheses

3F^2-12F+F-15=0

We add all the numbers together, and all the variables

3F^2-11F-15=0

a = 3; b = -11; c = -15;
Δ = b2-4ac
Δ = -112-4·3·(-15)
Δ = 301
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{-(-11)-\sqrt{301}}{2*3}=\frac{11-\sqrt{301}}{6}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{301}}{2*3}=\frac{11+\sqrt{301}}{6}
$