(h-1)*(3h-3)=2*(h-2)*(3h-3)-(h-2)*(h-1)

Solution for (h-1)*(3h-3)=2*(h-2)*(3h-3)-(h-2)*(h-1) equation:

(h-1)(3h-3)=2(h-2)(3h-3)-(h-2)(h-1)

We move all terms to the left:

(h-1)(3h-3)-(2(h-2)(3h-3)-(h-2)(h-1))=0

We multiply parentheses ..

(+3h^2-3h-3h+3)-(2(h-2)(3h-3)-(h-2)(h-1))=0


We calculate terms in parentheses: -(2(h-2)(3h-3)-(h-2)(h-1)), so:

2(h-2)(3h-3)-(h-2)(h-1)

We multiply parentheses ..

2(+3h^2-3h-6h+6)-(h-2)(h-1)

We multiply parentheses

6h^2-6h-12h-(h-2)(h-1)+12

We multiply parentheses ..

6h^2-(+h^2-1h-2h+2)-6h-12h+12

We add all the numbers together, and all the variables

6h^2-(+h^2-1h-2h+2)-18h+12

We get rid of parentheses

6h^2-h^2+1h+2h-18h-2+12

We add all the numbers together, and all the variables

5h^2-15h+10

Back to the equation:

-(5h^2-15h+10)


We get rid of parentheses

3h^2-5h^2-3h-3h+15h+3-10=0

We add all the numbers together, and all the variables

-2h^2+9h-7=0

a = -2; b = 9; c = -7;
Δ = b2-4ac
Δ = 92-4·(-2)·(-7)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-5}{2*-2}=\frac{-14}{-4}
=3+1/2
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+5}{2*-2}=\frac{-4}{-4}
=1
$