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(a+2)(a+3)=3

We move all terms to the left:

(a+2)(a+3)-(3)=0

We multiply parentheses ..

(+a^2+3a+2a+6)-3=0

We get rid of parentheses

a^2+3a+2a+6-3=0

We add all the numbers together, and all the variables

a^2+5a+3=0

a = 1; b = 5; c = +3;

Δ = b^{2}-4ac

Δ = 5^{2}-4·1·3

Δ = 13

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}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{13}}{2*1}=\frac{-5-\sqrt{13}}{2} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{13}}{2*1}=\frac{-5+\sqrt{13}}{2} $

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