How do you solve #11z^2-z=3# using the quadratic formula?

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Answer 1 · 2016-12-24T05:25:53

#z = [1 + sqrt(133)]/22 and [1 - sqrt(33)]/22#

#z~~ +-0.5697#

The general form of quadratic equation is #ax^2 +bx +c = 0#

and #x =(-b +-sqrt(b^2 - 4ac))/(2a)#

Here, the equation is #11z^2 - z = 3#

or, #" "11z^2 - z -3 = 0#

comparing this with #ax^2 +bx +c = 0#

we get #x =z, a = 11, b = - 1 and c = - 3#

Now, #z = =(-b +-sqrt(b^2 - 4ac))/(2a)# put values of a,b and c

or, #z =(-(-1) +-sqrt((-1)^2 - 4(11)(-3)))/(2(11))#

or, #z = =(1 +-sqrt(1 +132))/(22)#

or, #z = [1 + sqrt{133}]/22# and #z = (1 - sqrt{133})/22#