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)=0

We get rid of parentheses

k^2-11k-34=0

a = 1; b = -11; c = -34;
Δ = b2-4ac
Δ = -112-4·1·(-34)
Δ = 257
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-\sqrt{257}}{2*1}=\frac{11-\sqrt{257}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{257}}{2*1}=\frac{11+\sqrt{257}}{2}
$