3*l+(2-l)*l=(3-l)*(2-l)

Solution for 3*l+(2-l)*l=(3-l)*(2-l) equation:

3l+(2-l)*l=(3-l)(2-l)

We move all terms to the left:

3l+(2-l)*l-((3-l)(2-l))=0

We add all the numbers together, and all the variables

3l+(-1l+2)*l-((-1l+3)(-1l+2))=0

We multiply parentheses

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

We multiply parentheses ..

-1l^2-((+l^2-2l-3l+6))+3l+2l=0


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

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

We get rid of parentheses

l^2-2l-3l+6

We add all the numbers together, and all the variables

l^2-5l+6

Back to the equation:

-(l^2-5l+6)


We add all the numbers together, and all the variables

-1l^2+5l-(l^2-5l+6)=0

We get rid of parentheses

-1l^2-l^2+5l+5l-6=0

We add all the numbers together, and all the variables

-2l^2+10l-6=0

a = -2; b = 10; c = -6;
Δ = b2-4ac
Δ = 102-4·(-2)·(-6)
Δ = 52
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$l_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$l_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$l_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{13}}{2*-2}=\frac{-10-2\sqrt{13}}{-4}
$
$l_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{13}}{2*-2}=\frac{-10+2\sqrt{13}}{-4}
$