Solve k= k+15 divided by 3(k-1) use quadratic formula?

Solve k= k+15 divided by 3(k-1) use quadratic formula

Answer:
k = (k+15)/3*(k-1)

==> multiply both sides by 3(k-1) to get rid of fractions

3k(k-1) = k+15

==> expand the LHS

3k^2-3k = k+15

==> move everything to the LHS

3k^2-4k-15=0

actually this does factor into (3k+5)(k-3), but if you want to use the formula, just use:

a=3, b=-4 and c=-15

x = -(-4) +/- sqrt[(-4)^2-2*3*(-15)] / 2*3

x = 4 +/- sqrt(196) / 2*3

x = 4 +/- 14 / 6

==> factor out a 2 from top and bottom

x = 2 (2 +/- 7) / 2*3

==> cancel the twos to get

x = 2 +/- 7 / 3

which gives 2+7 / 3 and 2-7 / 3 which gives 3 and -5/3
multiply by 3(k-1)

3k^2 - 3k = k + 15

3k^2 - 4k + 15 = 0

quadratic equation

x = (-b +/- sqrt(b^2 - 4*a*c)) / (2a)

answers are

x1 = 0.66667 + 2.1344 i
x2 = 0.66667 - 2.1344 i

you're right math, missed a sign change on 15
multiply by 3(k-1)
3k(k-1)=k+15
3k^2-3k-k-15=0
3k^2-4k-15=0
k=3 or k=-1.67

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