How can i solve this equation?

Question

I will like to solve this equation using the quadratic formula:

here it is...

x^2 - 5x =0 (x^2 =x square)

I tried doing it but got stuck on the way..
Quadratic formula....Thanks in advance.

Please resolve the equation using the quadratic formula thanks

Jan W you just did the oposite thing..you didnt even bother reading my question

Answer 1

x = [ 5 ±√(25) ] / 2
x = 10 / 2 , x = 0
x = 5 , x = 0

Can also be solved using:-
(x) (x - 5) = 0
x = 0 , x = 5 as above.

Answer 2

well you don't have to do the quadratic formula to solve this one.

x(x - 5) = 0
x = 0, 5

using the formula: a = 1 b = -5 c = 0
x = 5 +- sqrt(25 - 4(1)(0))
-------------------------------------
2(1)

x = (5 +- sqrt(25)) / 2
x = (5 +- 5)/2
x = 10/2 = 5
x = 0/2 = 0

Answer 3

x^2 - 5x = 0 ==> x(x-5) = 0, therefore:
x1=0, x2=5.

Quadratic formula:

x1,2 = (-b +- SQRT(b^2-4ac))/2a

where:

SQRT means square root
a=1, b=5, c=0

Substitute into the equation:

x1,2 = ((5 +-SQRT(25-4*1*0))/2 ==>
x1,2 = ((5 +-SQRT(25))/2 ==>
x1,2 = (5 +-5)/2

x1 = (5+5)/2 = 10/2 = 5
x2 = (5-5)/2 = 0/2 = 0

Answer 4

Using the quadratic formula on this problem is totally lame. But since you asked, here is goes..
If ax^2 + bx + c = 0
then x = [-b +/- sqrt(b^2 - 4ac)]/(2a)
For your problem a = 1, b = -5, c = 0 so
b^2 - 4ac = 25 - 0 = 25
sqrt(b^2 - 4ac) = +/-5
then x = [5 +/- 5]/2
So x = 0 or x = 5

Answer 5

You do realize that you can solve it easier via factoring, right? That said, and mentioning that your "c" = 0,

x = (-b +/- SQRT(b^2 - 4ac)) / 2a ==

x = (-(-5) +/- SQRT((-5)^2 - 4(1)(0)) / 2(1) ==

x = (5 +/- SQRT(25 - 0)) / 2 ==

x = (5 +/- 5) / 2 ==

x = (0, 5)

Answer 6

x² - 5x =0
==> x(x - 5) = 0

x = 0

-or-

x - 5 = 0
==> x = 5

So: x = 0, 5

:)

Answer 7

x^2 - 5x =x * (x-5)
The condition
x * (x-5)=0
can be met when x=0 or when x=5