k(k+7)=11

Solution for k(k+7)=11 equation:

k(k+7)=11

We move all terms to the left:

k(k+7)-(11)=0

We multiply parentheses

k^2+7k-11=0

a = 1; b = 7; c = -11;
Δ = b2-4ac
Δ = 72-4·1·(-11)
Δ = 93
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{-(7)-\sqrt{93}}{2*1}=\frac{-7-\sqrt{93}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{93}}{2*1}=\frac{-7+\sqrt{93}}{2}
$