What are the roots of the quadratic equation below
Hint: Apply the quadratic formula
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
For a quadratic equation : \(ax^2+bx+c=0\)
solution:
\(x = {-15 \pm \sqrt{15^2-4*3*(-5)} \over 2(3)}\)\( = {-15 \pm \sqrt{165} \over 6}\)
