6k(k+3)=2(4k+5)

Solution for 6k(k+3)=2(4k+5) equation:

6k(k+3)=2(4k+5)

We move all terms to the left:

6k(k+3)-(2(4k+5))=0

We multiply parentheses

6k^2+18k-(2(4k+5))=0


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

2(4k+5)

We multiply parentheses

8k+10

Back to the equation:

-(8k+10)


We get rid of parentheses

6k^2+18k-8k-10=0

We add all the numbers together, and all the variables

6k^2+10k-10=0

a = 6; b = 10; c = -10;
Δ = b2-4ac
Δ = 102-4·6·(-10)
Δ = 340
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:

$\sqrt{\Delta}=\sqrt{340}=\sqrt{4*85}=\sqrt{4}*\sqrt{85}=2\sqrt{85}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{85}}{2*6}=\frac{-10-2\sqrt{85}}{12}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{85}}{2*6}=\frac{-10+2\sqrt{85}}{12}
$