# How do you find the value of the discriminant and determine the nature of the roots -x^2-9=6x?

Apr 17, 2018

$D$=0 and the roots are real and equal.

#### Explanation:

We have, ${x}^{2} + 6 x + 9$=0 [On arrangement]
Thus, $a$=1,$b$=6 and $c$=9
Thus,$D = {b}^{2} - 4 a c$
=${\left(6\right)}^{2} - 4 \times 1 \times 9$
=$36 - 36$=0
Thus,the roots are real and equal to $x = - 3 , - 3$