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3x(x-2)+1=3x+5
We move all terms to the left:
3x(x-2)+1-(3x+5)=0
We multiply parentheses
3x^2-6x-(3x+5)+1=0
We get rid of parentheses
3x^2-6x-3x-5+1=0
We add all the numbers together, and all the variables
3x^2-9x-4=0
a = 3; b = -9; c = -4;
Δ = b2-4ac
Δ = -92-4·3·(-4)
Δ = 129
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-\sqrt{129}}{2*3}=\frac{9-\sqrt{129}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{129}}{2*3}=\frac{9+\sqrt{129}}{6} $
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