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(2x+3)(3x+1)=3
We move all terms to the left:
(2x+3)(3x+1)-(3)=0
We multiply parentheses ..
(+6x^2+2x+9x+3)-3=0
We get rid of parentheses
6x^2+2x+9x+3-3=0
We add all the numbers together, and all the variables
6x^2+11x=0
a = 6; b = 11; c = 0;
Δ = b2-4ac
Δ = 112-4·6·0
Δ = 121
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{121}=11$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-11}{2*6}=\frac{-22}{12} =-1+5/6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+11}{2*6}=\frac{0}{12} =0 $
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