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(3-2x)(2-3x)=8x+6
We move all terms to the left:
(3-2x)(2-3x)-(8x+6)=0
We add all the numbers together, and all the variables
(-2x+3)(-3x+2)-(8x+6)=0
We get rid of parentheses
(-2x+3)(-3x+2)-8x-6=0
We multiply parentheses ..
(+6x^2-4x-9x+6)-8x-6=0
We get rid of parentheses
6x^2-4x-9x-8x+6-6=0
We add all the numbers together, and all the variables
6x^2-21x=0
a = 6; b = -21; c = 0;
Δ = b2-4ac
Δ = -212-4·6·0
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{441}=21$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-21}{2*6}=\frac{0}{12} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+21}{2*6}=\frac{42}{12} =3+1/2 $
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