Question

How do you find the zeros, if any, of `y= -3x^2 +1 3x+31`using the quadratic formula?

Answer 1

Zeros of `y(x)=-3x^2+13x+31` are `13/6+sqrt541/6` and `13/6-sqrt541/6`.

Explanation:

According to quadratic formula, zeros of `y(x)=ax^2+bx+c` are given by `x=(-b+-sqrt(b^2-4ac))/(2a)`.

Hence, zeros of `y(x)=-3x^2+13x+31` are given by

`x=(-13+-sqrt(13^2-4×(-3)×31))/(2×(-3)`

= `(-13+-sqrt(169+372))/(-6)`

= `(-13+-sqrt541)/(-6)`

= `13/6+-sqrt541/6`

Hence zeros of `y(x)=-3x^2+13x+31` are `13/6+sqrt541/6` and `13/6-sqrt541/6`.