hard quadratic equations!!?

Question

I dont understand theses questions!!

7. x^2 + 7x + 9 = 0

8. x^2 + + 6x = 0

9. 15x^2 – 7x – 2 = 0

10. x2 – 9x + 5 = 0

supposed to solve the quadratic equations using the quadratic formula!

Answer 1

So then solve using the quadratic. I'll do 7.

(-7+-sqrt49-36)/2
(-7+-sqrt13)/2
x=-7/2+sqrt13/2
x=-7/2-sqrt13/2

Remember the formula is:
(-b+-sqrt b^2-4ac)/2a

Answer 2

7. x^2 + 7x + 9 = 0

For ax2 + bx + c = 0, the value of x is given by:

-b + or - sqrt (b^2 - 4ac) / 2a (Note that, for the Formula to work, you must have "(quadratic) = 0". Note also that the "2a" at the bottom of the Formula is underneath everything above, not just the square root. And don't forget that it's a "2a" under there, not just a "2")

So, here we have a = 1, b = 7 and c = 9

x1 = -7 + sqrt(49 - 36) / 2 = (-7 + sqrt 13) / 2 and

x2 = (- 7 - sqrt13) / 2

Try the other problems similarly.

Answer 3

What is there to not understand? Just determine a, b and c, substitute into the formula and solve. For example in #9, a= 15, b = -7 and c= -2.
x = [-(-7) +/- sqrt((-7)^2- 4*15*-2 ) ]/(2*15)
x= 7+/-sqrt(169)/30, x= 7 +/- 13/30 , x=2/3,-1/5

Answer 4

If you have an equation
ax^2+bx+c=0
the x´s values are given by
x=((-b+-sqrt(b^2-4ac))/2a
Applying this formula for the first
x=((-7+-sqrt(49-36)/2 =-7/2+-1/2sqrt(13)
2)taking x out x((x+6)=0 so x=0 and x=-6
9 and 10 can be solved using the formula

Answer 5

OMG im doing these too!!

okay im still practicing, but I am going to try urs out anyways :)

7) x(squared)+7x+9=0

x(squared)+6x+1x+9=0

x(x+6x+1x+9)=0

(x+6x)(1x+9)=0

(3,1/9)

Answer 6

you have to slove for X.
so use the quadratic formula: [-b +/- (b^2 +4ac)^1/2 ] / 2a

for #7- a=1
b=7
c=9
and voila!

Answer 7

what is the question are you suposed to factor, solve? that would be helpfull to know!