a+(a)2+1=41

Solution for a+(a)2+1=41 equation:

a+(a)2+1=41

We move all terms to the left:

a+(a)2+1-(41)=0

We add all the numbers together, and all the variables

a^2+a-40=0

a = 1; b = 1; c = -40;
Δ = b2-4ac
Δ = 12-4·1·(-40)
Δ = 161
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{-(1)-\sqrt{161}}{2*1}=\frac{-1-\sqrt{161}}{2}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{161}}{2*1}=\frac{-1+\sqrt{161}}{2}
$