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(a+1)(3a+2)=0
We multiply parentheses ..
(+3a^2+2a+3a+2)=0
We get rid of parentheses
3a^2+2a+3a+2=0
We add all the numbers together, and all the variables
3a^2+5a+2=0
a = 3; b = 5; c = +2;
Δ = b2-4ac
Δ = 52-4·3·2
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1}=1$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-1}{2*3}=\frac{-6}{6} =-1 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+1}{2*3}=\frac{-4}{6} =-2/3 $
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