F(x)=(x-6)(2x+10)

Solution for F(x)=(x-6)(2x+10) equation:

(F)=(F-6)(2F+10)

We move all terms to the left:

(F)-((F-6)(2F+10))=0

We multiply parentheses ..

-((+2F^2+10F-12F-60))+F=0


We calculate terms in parentheses: -((+2F^2+10F-12F-60)), so:

(+2F^2+10F-12F-60)

We get rid of parentheses

2F^2+10F-12F-60

We add all the numbers together, and all the variables

2F^2-2F-60

Back to the equation:

-(2F^2-2F-60)


We add all the numbers together, and all the variables

F-(2F^2-2F-60)=0

We get rid of parentheses

-2F^2+F+2F+60=0

We add all the numbers together, and all the variables

-2F^2+3F+60=0

a = -2; b = 3; c = +60;
Δ = b2-4ac
Δ = 32-4·(-2)·60
Δ = 489
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{-(3)-\sqrt{489}}{2*-2}=\frac{-3-\sqrt{489}}{-4}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{489}}{2*-2}=\frac{-3+\sqrt{489}}{-4}
$