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

Dec 24, 2016

$z = \frac{1 + \sqrt{133}}{22} \mathmr{and} \frac{1 - \sqrt{33}}{22}$

$z \approx \pm 0.5697$

#### Explanation:

The general form of quadratic equation is $a {x}^{2} + b x + c = 0$

and $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Here, the equation is $11 {z}^{2} - z = 3$

or, $\text{ } 11 {z}^{2} - z - 3 = 0$

comparing this with $a {x}^{2} + b x + c = 0$

we get $x = z , a = 11 , b = - 1 \mathmr{and} c = - 3$

Now, $z = = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$ put values of a,b and c

or, $z = \frac{- \left(- 1\right) \pm \sqrt{{\left(- 1\right)}^{2} - 4 \left(11\right) \left(- 3\right)}}{2 \left(11\right)}$

or, $z = = \frac{1 \pm \sqrt{1 + 132}}{22}$

or, $z = \frac{1 + \sqrt{133}}{22}$ and $z = \frac{1 - \sqrt{133}}{22}$