How do you find the zeros, if any, of #y= -x^2 -5x+2#using the quadratic formula?

1 Answer
Apr 7, 2017

Answer:

Zeros are # x ~~ -5.37 , x~~ 0.37#

Explanation:

Quadratic equation is #y=ax^2 +bx+c#

#y= -x^2 -5x +2 ; a = -1 , b= -5 , c=2 ; D= b^2-4*a*c = 25 +8 =33 (+)#
Since D is positve, roots are real.

Roots are # x= -b/(2a)+- sqrt(b^2-4ac)/(2a) or x = 5/-2 +- sqrt 33/-2 or x= -2.5 +- (-2.87) :. x= -2.5 -2.87 ~~ -5.37 or x~~ 0.37# [Ans]