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

$x = 3 \pm i \sqrt{\frac{3}{5}}$
For using quadratic formula, the given expression has to be put in quadratic form , that is $5 {x}^{2} - 30 x + 48 = 0$. Comparing with the quadratic formula a= 5, b= -30 and c= 48. Expression ${b}^{2} - 4 a c < 0$ hence the equation would have imaginary roots. The roots would be $x = 3 \pm i \sqrt{\frac{3}{5}}$