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

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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} $

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