k^2+16k+68=5

Solution for k^2+16k+68=5 equation:

k^2+16k+68=5

We move all terms to the left:

k^2+16k+68-(5)=0

We add all the numbers together, and all the variables

k^2+16k+63=0

a = 1; b = 16; c = +63;
Δ = b2-4ac
Δ = 162-4·1·63
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2}{2*1}=\frac{-18}{2}
=-9
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2}{2*1}=\frac{-14}{2}
=-7
$