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

Solution for (k-3)(2k+3)=0 equation:

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

We multiply parentheses ..

(+2k^2+3k-6k-9)=0

We get rid of parentheses

2k^2+3k-6k-9=0

We add all the numbers together, and all the variables

2k^2-3k-9=0

a = 2; b = -3; c = -9;
Δ = b2-4ac
Δ = -32-4·2·(-9)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-9}{2*2}=\frac{-6}{4}
=-1+1/2
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+9}{2*2}=\frac{12}{4}
=3
$