Could you check and see if my answers are right?

Question

Using the quadratic equation to solve:

a^2 - 4a - 21 = 0
x = 8.123 & -13.175

m^2 + 4m - 5 =0
x=-1 & -6

x^2 = 9x =0
x= -13.5 & -4.5

x^2 + 2x = 0
x= -1 & -4

-x^2 = 6x -9=0
x=8.74 & -27

This is optional, but it would help me. If I'm wrong can you show your steps so I can see my mistake. Thanks for the help!

Josh, I need the work for the last 3!

Answer 1

If you want to check your answers, there are a number of quadratic equation solvers on the web, such as:
http://www.math.com/students/calculators/source/quadratic.htm

Of course, you can always check to see if your answer is right by substituting each solution for x and doing the math.

Answer 2

noooo...
a^2-4a-21=0 (divide the trinomial into two binomials)
(a-7)=0 (a+3)=0
+7 │+7 -3│-3
a=7 a=-3

m^2+4m-5=0
(m+5) (m-1)=0 (The two numbers must add up to the middle interger in the trinomial and multiply to the third)
m=-5 m=1


x=3, -3

x=2, 0

x=3

message me if u need the work for the last three

Answer 3

I'm sorry, I hate math , just the sight of numbers makes me feel funny. sorry

Answer 4

well of course it optional. duh der.