k(2k+3)=3

Solution for k(2k+3)=3 equation:

k(2k+3)=3

We move all terms to the left:

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

We multiply parentheses

2k^2+3k-3=0

a = 2; b = 3; c = -3;
Δ = b2-4ac
Δ = 32-4·2·(-3)
Δ = 33
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{33}}{2*2}=\frac{-3-\sqrt{33}}{4}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{33}}{2*2}=\frac{-3+\sqrt{33}}{4}
$