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a+a2+a+3=39
We move all terms to the left:
a+a2+a+3-(39)=0
We add all the numbers together, and all the variables
a^2+2a-36=0
a = 1; b = 2; c = -36;
Δ = b2-4ac
Δ = 22-4·1·(-36)
Δ = 148
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}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{148}=\sqrt{4*37}=\sqrt{4}*\sqrt{37}=2\sqrt{37}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{37}}{2*1}=\frac{-2-2\sqrt{37}}{2} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{37}}{2*1}=\frac{-2+2\sqrt{37}}{2} $
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