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(3x-1)(x+2)-(2x-3)(x-2)=2x+3
We move all terms to the left:
(3x-1)(x+2)-(2x-3)(x-2)-(2x+3)=0
We get rid of parentheses
(3x-1)(x+2)-(2x-3)(x-2)-2x-3=0
We multiply parentheses ..
(+3x^2+6x-1x-2)-(2x-3)(x-2)-2x-3=0
We add all the numbers together, and all the variables
(+3x^2+6x-1x-2)-2x-(2x-3)(x-2)-3=0
We get rid of parentheses
3x^2+6x-1x-2x-(2x-3)(x-2)-2-3=0
We multiply parentheses ..
3x^2-(+2x^2-4x-3x+6)+6x-1x-2x-2-3=0
We add all the numbers together, and all the variables
3x^2-(+2x^2-4x-3x+6)+3x-5=0
We get rid of parentheses
3x^2-2x^2+4x+3x+3x-6-5=0
We add all the numbers together, and all the variables
x^2+10x-11=0
a = 1; b = 10; c = -11;
Δ = b2-4ac
Δ = 102-4·1·(-11)
Δ = 144
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{144}=12$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-12}{2*1}=\frac{-22}{2} =-11 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+12}{2*1}=\frac{2}{2} =1 $
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