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 $z = (1 - sqrt{133})/22$

How do you solve 11z^2-z=311z^2-z=3 using the quadratic formula?