15k(k+2)=2k-5

Solution for 15k(k+2)=2k-5 equation:

15k(k+2)=2k-5

We move all terms to the left:

15k(k+2)-(2k-5)=0

We multiply parentheses

15k^2+30k-(2k-5)=0

We get rid of parentheses

15k^2+30k-2k+5=0

We add all the numbers together, and all the variables

15k^2+28k+5=0

a = 15; b = 28; c = +5;
Δ = b2-4ac
Δ = 282-4·15·5
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-22}{2*15}=\frac{-50}{30}
=-1+2/3
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+22}{2*15}=\frac{-6}{30}
=-1/5
$