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

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

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

We move all terms to the left:

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

We multiply parentheses ..

-((+12F^2-15F+8F-10))+F=0


We calculate terms in parentheses: -((+12F^2-15F+8F-10)), so:

(+12F^2-15F+8F-10)

We get rid of parentheses

12F^2-15F+8F-10

We add all the numbers together, and all the variables

12F^2-7F-10

Back to the equation:

-(12F^2-7F-10)


We add all the numbers together, and all the variables

F-(12F^2-7F-10)=0

We get rid of parentheses

-12F^2+F+7F+10=0

We add all the numbers together, and all the variables

-12F^2+8F+10=0

a = -12; b = 8; c = +10;
Δ = b2-4ac
Δ = 82-4·(-12)·10
Δ = 544
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:

$\sqrt{\Delta}=\sqrt{544}=\sqrt{16*34}=\sqrt{16}*\sqrt{34}=4\sqrt{34}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{34}}{2*-12}=\frac{-8-4\sqrt{34}}{-24}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{34}}{2*-12}=\frac{-8+4\sqrt{34}}{-24}
$