3k(k+1)+11=2(4+5k)+3

Solution for 3k(k+1)+11=2(4+5k)+3 equation:

3k(k+1)+11=2(4+5k)+3

We move all terms to the left:

3k(k+1)+11-(2(4+5k)+3)=0

We add all the numbers together, and all the variables

3k(k+1)-(2(5k+4)+3)+11=0

We multiply parentheses

3k^2+3k-(2(5k+4)+3)+11=0


We calculate terms in parentheses: -(2(5k+4)+3), so:

2(5k+4)+3

We multiply parentheses

10k+8+3

We add all the numbers together, and all the variables

10k+11

Back to the equation:

-(10k+11)


We get rid of parentheses

3k^2+3k-10k-11+11=0

We add all the numbers together, and all the variables

3k^2-7k=0

a = 3; b = -7; c = 0;
Δ = b2-4ac
Δ = -72-4·3·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-7}{2*3}=\frac{0}{6}
=0
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+7}{2*3}=\frac{14}{6}
=2+1/3
$