3k+6=(k+4)(k+6)/4

Solution for 3k+6=(k+4)(k+6)/4 equation:

3k+6=(k+4)(k+6)/4

We move all terms to the left:

3k+6-((k+4)(k+6)/4)=0

We multiply parentheses ..

-((+k^2+6k+4k+24)/4)+3k+6=0

We multiply all the terms by the denominator


-((+k^2+6k+4k+24)+3k*4)+6*4)=0


We calculate terms in parentheses: -((+k^2+6k+4k+24)+3k*4), so:

(+k^2+6k+4k+24)+3k*4

Wy multiply elements

(+k^2+6k+4k+24)+12k

We get rid of parentheses

k^2+6k+4k+12k+24

We add all the numbers together, and all the variables

k^2+22k+24

Back to the equation:

-(k^2+22k+24)


We add all the numbers together, and all the variables

-(k^2+22k+24)=0

We get rid of parentheses

-k^2-22k-24=0

We add all the numbers together, and all the variables

-1k^2-22k-24=0

a = -1; b = -22; c = -24;
Δ = b2-4ac
Δ = -222-4·(-1)·(-24)
Δ = 388
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{388}=\sqrt{4*97}=\sqrt{4}*\sqrt{97}=2\sqrt{97}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{97}}{2*-1}=\frac{22-2\sqrt{97}}{-2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{97}}{2*-1}=\frac{22+2\sqrt{97}}{-2}
$