H(x)=(3-x)(2-x)

Solution for H(x)=(3-x)(2-x) equation:

(H)=(3-H)(2-H)

We move all terms to the left:

(H)-((3-H)(2-H))=0

We add all the numbers together, and all the variables

H-((-1H+3)(-1H+2))=0

We multiply parentheses ..

-((+H^2-2H-3H+6))+H=0


We calculate terms in parentheses: -((+H^2-2H-3H+6)), so:

(+H^2-2H-3H+6)

We get rid of parentheses

H^2-2H-3H+6

We add all the numbers together, and all the variables

H^2-5H+6

Back to the equation:

-(H^2-5H+6)


We add all the numbers together, and all the variables

H-(H^2-5H+6)=0

We get rid of parentheses

-H^2+H+5H-6=0

We add all the numbers together, and all the variables

-1H^2+6H-6=0

a = -1; b = 6; c = -6;
Δ = b2-4ac
Δ = 62-4·(-1)·(-6)
Δ = 12
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{3}}{2*-1}=\frac{-6-2\sqrt{3}}{-2}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{3}}{2*-1}=\frac{-6+2\sqrt{3}}{-2}
$