5 units left and 2 units down
Keywords
quadratic equation, discriminant, complex roots, real roots
we know that
The formula to calculate the <u>roots</u> of the <u>quadratic equation</u> of the form
is equal to

where
The <u>discriminant</u> of the <u>quadratic equation</u> is equal to

if
----> the <u>quadratic equation</u> has two <u>real roots</u>
if
----> the <u>quadratic equation</u> has one <u>real root</u>
if
----> the <u>quadratic equation</u> has two <u>complex roots</u>
in this problem we have that
the <u>discriminant</u> is equal to 
so
the <u>quadratic equation</u> has two <u>complex roots</u>
therefore
the answer is the option A
There are two complex roots
3(1+2)
3+6
9 ( the answer is nine)
first you divide six and two then use the distributive property to multiply three and one and three and two, then add three and six to get nine.
So there are 41 two pointers and three pointers.
there are 21 two pointers and 20 three pointers