k(k+3)=19

Solution for k(k+3)=19 equation:

k(k+3)=19

We move all terms to the left:

k(k+3)-(19)=0

We multiply parentheses

k^2+3k-19=0

a = 1; b = 3; c = -19;
Δ = b2-4ac
Δ = 32-4·1·(-19)
Δ = 85
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{-(3)-\sqrt{85}}{2*1}=\frac{-3-\sqrt{85}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{85}}{2*1}=\frac{-3+\sqrt{85}}{2}
$