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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 = b2-4ac
Δ = 52-4·1·3
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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