10k(k+3)=3k-5

Solution for 10k(k+3)=3k-5 equation:

10k(k+3)=3k-5

We move all terms to the left:

10k(k+3)-(3k-5)=0

We multiply parentheses

10k^2+30k-(3k-5)=0

We get rid of parentheses

10k^2+30k-3k+5=0

We add all the numbers together, and all the variables

10k^2+27k+5=0

a = 10; b = 27; c = +5;
Δ = b2-4ac
Δ = 272-4·10·5
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-23}{2*10}=\frac{-50}{20}
=-2+1/2
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+23}{2*10}=\frac{-4}{20}
=-1/5
$