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(2x+1)(3x+1)=0
We multiply parentheses ..
(+6x^2+2x+3x+1)=0
We get rid of parentheses
6x^2+2x+3x+1=0
We add all the numbers together, and all the variables
6x^2+5x+1=0
a = 6; b = 5; c = +1;
Δ = b2-4ac
Δ = 52-4·6·1
Δ = 1
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{1}=1$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-1}{2*6}=\frac{-6}{12} =-1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+1}{2*6}=\frac{-4}{12} =-1/3 $
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