How do you solve #-7= - 3z ^ { 2} + 9z# using the quadratic formula?

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Answer 1 · 2017-11-21T22:30:19

#z = (9 +- sqrt(165))/6#

#-7 = -3z^2 + 9z#

The first thing we want to do is set one side to be 0, so let's move everything to the left side of the equation.

#3z^2 - 9z - 7 = 0#

This equation is now written in standard quadratic form, or #ax^2 + bx + c#.

Now we can use the quadratic formula. The quadratic formula is #(-b +- sqrt(b^2 - 4ac))/(2a)#

We know that #a = 3#, #b = -9#, and #c = -7#, so let's plug the numbers into the quadratic formula.

#z = (-(-9) +- sqrt((-9)^2 - 4(3)(-7)))/(2(3))#

#z = (9 +- sqrt(81-4(-21)))/6#

#z = (9 +- sqrt(81+84))/6#

#z = (9 +- sqrt(165))/6#