a^2+11a+29=1

Solution for a^2+11a+29=1 equation:

a^2+11a+29=1

We move all terms to the left:

a^2+11a+29-(1)=0

We add all the numbers together, and all the variables

a^2+11a+28=0

a = 1; b = 11; c = +28;
Δ = b2-4ac
Δ = 112-4·1·28
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-3}{2*1}=\frac{-14}{2}
=-7
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+3}{2*1}=\frac{-8}{2}
=-4
$