6(6k-8)-20=11(2k-3)3k

Solution for 6(6k-8)-20=11(2k-3)3k equation:

6(6k-8)-20=11(2k-3)3k

We move all terms to the left:

6(6k-8)-20-(11(2k-3)3k)=0

We multiply parentheses

36k-(11(2k-3)3k)-48-20=0


We calculate terms in parentheses: -(11(2k-3)3k), so:

11(2k-3)3k

We multiply parentheses

66k^2-99k

Back to the equation:

-(66k^2-99k)


We add all the numbers together, and all the variables

36k-(66k^2-99k)-68=0

We get rid of parentheses

-66k^2+36k+99k-68=0

We add all the numbers together, and all the variables

-66k^2+135k-68=0

a = -66; b = 135; c = -68;
Δ = b2-4ac
Δ = 1352-4·(-66)·(-68)
Δ = 273
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(135)-\sqrt{273}}{2*-66}=\frac{-135-\sqrt{273}}{-132}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(135)+\sqrt{273}}{2*-66}=\frac{-135+\sqrt{273}}{-132}
$