k^2+11k+34=6

Solution for k^2+11k+34=6 equation:

k^2+11k+34=6

We move all terms to the left:

k^2+11k+34-(6)=0

We add all the numbers together, and all the variables

k^2+11k+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:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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