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

1 Answer
Aug 25, 2016

Answer:

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#.