# How do you find the roots, real and imaginary, of y= x^2 - 121  using the quadratic formula?

Jun 8, 2018

The two solutions are $11$ and $- 11$

#### Explanation:

Compare your equation, ${x}^{2} - 121$, with the generic equation $a {x}^{2} + b x + c$. Your equation is obtained by choosing $a = 1$, $b = 0$, $c = - 121$.

The quadratic formula states that the solution of a quadratic equation are

${x}_{1 , 2} = \setminus \frac{- b \setminus \pm \setminus \sqrt{{b}^{2} - 4 a c}}{2 a}$

${x}_{1 , 2} = \setminus \frac{0 \setminus \pm \setminus \sqrt{{0}^{2} - 4 \cdot 1 \cdot \left(- 121\right)}}{2 \cdot 1} = \setminus \pm \setminus \frac{\sqrt{484}}{2} = \setminus \pm \setminus \frac{22}{2} = \setminus \pm 11$
So, the two solutions are $11$ and $- 11$