F(x)=3(4x-5)(x-2)

Solution for F(x)=3(4x-5)(x-2) equation:

(F)=3(4F-5)(F-2)

We move all terms to the left:

(F)-(3(4F-5)(F-2))=0

We multiply parentheses ..

-(3(+4F^2-8F-5F+10))+F=0


We calculate terms in parentheses: -(3(+4F^2-8F-5F+10)), so:

3(+4F^2-8F-5F+10)

We multiply parentheses

12F^2-24F-15F+30

We add all the numbers together, and all the variables

12F^2-39F+30

Back to the equation:

-(12F^2-39F+30)


We add all the numbers together, and all the variables

F-(12F^2-39F+30)=0

We get rid of parentheses

-12F^2+F+39F-30=0

We add all the numbers together, and all the variables

-12F^2+40F-30=0

a = -12; b = 40; c = -30;
Δ = b2-4ac
Δ = 402-4·(-12)·(-30)
Δ = 160
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-4\sqrt{10}}{2*-12}=\frac{-40-4\sqrt{10}}{-24}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+4\sqrt{10}}{2*-12}=\frac{-40+4\sqrt{10}}{-24}
$