4k(2k+3)=2k(3k+2)

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

4k(2k+3)=2k(3k+2)

We move all terms to the left:

4k(2k+3)-(2k(3k+2))=0

We multiply parentheses

8k^2+12k-(2k(3k+2))=0


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

2k(3k+2)

We multiply parentheses

6k^2+4k

Back to the equation:

-(6k^2+4k)


We get rid of parentheses

8k^2-6k^2+12k-4k=0

We add all the numbers together, and all the variables

2k^2+8k=0

a = 2; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·2·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8}{2*2}=\frac{-16}{4}
=-4
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*2}=\frac{0}{4}
=0
$