(3k^2-5k)=(-6+2k^2)

Solution for (3k^2-5k)=(-6+2k^2) equation:

(3k^2-5k)=(-6+2k^2)

We move all terms to the left:

(3k^2-5k)-((-6+2k^2))=0

We get rid of parentheses

-((-6+2k^2))+3k^2-5k=0


We calculate terms in parentheses: -((-6+2k^2)), so:

(-6+2k^2)

We get rid of parentheses

2k^2-6

Back to the equation:

-(2k^2-6)


We add all the numbers together, and all the variables

3k^2-5k-(2k^2-6)=0

We get rid of parentheses

3k^2-2k^2-5k+6=0

We add all the numbers together, and all the variables

k^2-5k+6=0

a = 1; b = -5; c = +6;
Δ = b2-4ac
Δ = -52-4·1·6
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-1}{2*1}=\frac{4}{2}
=2
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+1}{2*1}=\frac{6}{2}
=3
$