# How do you find the roots, real and imaginary, of y=3(x - 6)^2-x + 3 using the quadratic formula?

Feb 13, 2016

Roots are $\frac{37 + \sqrt{37}}{6}$ and $\frac{37 - \sqrt{37}}{6}$, both real but irrational

#### Explanation:

Converting the expression to its general form

3(x−6)^2−x+3=3(x^2-12x+36)-x+3

or $3 {x}^{2} - 37 x + 111$

Quadratic formula tell that the root of this are given by

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

As $a = 3 , b = - 37 \mathmr{and} c = 111$, roots of given equation are given by

$\frac{37 \pm \sqrt{1369 - 1332}}{6}$ i.e. $\frac{37 \pm \sqrt{37}}{6}$

i.e. $\frac{37 + \sqrt{37}}{6}$ and $\frac{37 - \sqrt{37}}{6}$, both real but irrational