2(h+1)=(h+1)(h+1)

Solution for 2(h+1)=(h+1)(h+1) equation:

2(h+1)=(h+1)(h+1)

We move all terms to the left:

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

We multiply parentheses

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

We multiply parentheses ..

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


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

(+h^2+h+h+1)

We get rid of parentheses

h^2+h+h+1

We add all the numbers together, and all the variables

h^2+2h+1

Back to the equation:

-(h^2+2h+1)


We add all the numbers together, and all the variables

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

We get rid of parentheses

-h^2+2h-2h-1+2=0

We add all the numbers together, and all the variables

-1h^2+1=0

a = -1; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-1)·1
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2}{2*-1}=\frac{-2}{-2}
=1
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2}{2*-1}=\frac{2}{-2}
=-1
$