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 = 929
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{-(27)-\sqrt{929}}{2*10}=\frac{-27-\sqrt{929}}{20}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+\sqrt{929}}{2*10}=\frac{-27+\sqrt{929}}{20}
$