k(2k-3)=19

Solution for k(2k-3)=19 equation:

k(2k-3)=19

We move all terms to the left:

k(2k-3)-(19)=0

We multiply parentheses

2k^2-3k-19=0

a = 2; b = -3; c = -19;
Δ = b2-4ac
Δ = -32-4·2·(-19)
Δ = 161
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{161}}{2*2}=\frac{3-\sqrt{161}}{4}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{161}}{2*2}=\frac{3+\sqrt{161}}{4}
$