# How do you find the roots, real and imaginary, of y=-(x-3)^2+4x^2-x-5  using the quadratic formula?

Apr 4, 2016

Roots are real (irrational) numbers $\frac{- 5 + \sqrt{193}}{6}$ and $\frac{- 5 - \sqrt{193}}{6}$

#### Explanation:

Simplifying $y = - {\left(x - 3\right)}^{2} + 4 {x}^{2} - x - 5 = - \left({x}^{2} - 6 x + 9\right) + 4 {x}^{2} - x - 5$

or $y = 3 {x}^{2} + 5 x - 14$

Its root will be given by the equation $3 {x}^{2} + 5 x - 14 = 0$.

Using quadratic formula $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

these are $\frac{- 5 \pm \sqrt{{5}^{2} - 4 \cdot 3 \cdot \left(- 14\right)}}{2 \cdot 3} = \frac{- 5 \pm \sqrt{25 + 168}}{6} = \frac{- 5 \pm \sqrt{193}}{6}$

Hence roots are irrational numbers $\frac{- 5 + \sqrt{193}}{6}$ and $\frac{- 5 - \sqrt{193}}{6}$