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(2x+1)(x-1)=(3x+1)(x-2)
We move all terms to the left:
(2x+1)(x-1)-((3x+1)(x-2))=0
We multiply parentheses ..
(+2x^2-2x+x-1)-((3x+1)(x-2))=0
We calculate terms in parentheses: -((3x+1)(x-2)), so:We get rid of parentheses
(3x+1)(x-2)
We multiply parentheses ..
(+3x^2-6x+x-2)
We get rid of parentheses
3x^2-6x+x-2
We add all the numbers together, and all the variables
3x^2-5x-2
Back to the equation:
-(3x^2-5x-2)
2x^2-3x^2-2x+x+5x-1+2=0
We add all the numbers together, and all the variables
-1x^2+4x+1=0
a = -1; b = 4; c = +1;
Δ = b2-4ac
Δ = 42-4·(-1)·1
Δ = 20
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{5}}{2*-1}=\frac{-4-2\sqrt{5}}{-2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{5}}{2*-1}=\frac{-4+2\sqrt{5}}{-2} $
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