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(x+3)(3x+1)=3
We move all terms to the left:
(x+3)(3x+1)-(3)=0
We multiply parentheses ..
(+3x^2+x+9x+3)-3=0
We get rid of parentheses
3x^2+x+9x+3-3=0
We add all the numbers together, and all the variables
3x^2+10x=0
a = 3; b = 10; c = 0;
Δ = b2-4ac
Δ = 102-4·3·0
Δ = 100
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{100}=10$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10}{2*3}=\frac{-20}{6} =-3+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10}{2*3}=\frac{0}{6} =0 $
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