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

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

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

We move all terms to the left:

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

We multiply parentheses ..

-((+3F^2-15F+F-5))+F=0


We calculate terms in parentheses: -((+3F^2-15F+F-5)), so:

(+3F^2-15F+F-5)

We get rid of parentheses

3F^2-15F+F-5

We add all the numbers together, and all the variables

3F^2-14F-5

Back to the equation:

-(3F^2-14F-5)


We add all the numbers together, and all the variables

F-(3F^2-14F-5)=0

We get rid of parentheses

-3F^2+F+14F+5=0

We add all the numbers together, and all the variables

-3F^2+15F+5=0

a = -3; b = 15; c = +5;
Δ = b2-4ac
Δ = 152-4·(-3)·5
Δ = 285
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{-(15)-\sqrt{285}}{2*-3}=\frac{-15-\sqrt{285}}{-6}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{285}}{2*-3}=\frac{-15+\sqrt{285}}{-6}
$