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

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Answer 1 · 2016-08-25T22:56:46

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

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 $13/6-sqrt541/6$ .

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