SOLUTION: Can you solve the equation using quadratic formula x^2-5x-24=0 can you please show me the steps so I can do the next one on my own.

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Question 94106: Can you solve the equation using quadratic formula x^2-5x-24=0 can you please show me the steps so I can do the next one on my own.
Found 2 solutions by stanbon, praseenakos@yahoo.com:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
solve the equation using quadratic formula x^2-5x-24=0
x = [5+-sqrt(5^2-4*1*-24)]/(2*1)
x = [5 +- sqrt(121)]/2
x = [5 +- 11]/2
x = (5+11)/2 = 8 or x = (5-11)/2 = -3
----------
The formula is:
x = [-b +- sqrt ( b^2-4ac)]/(2a)
=================
Cheers,
Stan H.

Answer by praseenakos@yahoo.com(507) About Me  (Show Source):
You can put this solution on YOUR website!
Question
Can you solve the equation using quadratic formula x^2-5x-24=0 can you please show me the steps so I can do the next one on my own.

Answer:

The standard for of a quadratic equation is,

+a+x%5E2+%2B+bx+%2B+c+=+0+ _________________________(1)

For this standard equation, the solution is given by( Quadratic formula)

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

Given equation is,

+x%5E2+-+5x+-+24++=+0+ ________________(2)

Comparing this equation with the standard equation, you have the values,

+a+=+1%2C+b+=+-5+and+c+=+-24

Now substitute these values in the quadratic formula.....

==> x+=+%28-%28-5%29+%2B-+sqrt%28+%28-5%29%5E2-4%2A1%2A%28-24%29+%29%29%2F%282%2A1%29+


==> ==> x+=+%28+5++%2B-+sqrt%28+25+%2B+96+%29%29%2F+2+



==> x+=+%28+5++%2B-+sqrt%28+121%29%29%2F+2+


==> x+=++%285++%2B-+11%29%2F+2++


==> either x+=++%285++%2B+11%29%2F+2++ or x+=++%285++-+11%29%2F+2++


==> either x+=+16%2F+2++ or x+=+-6%2F+2++



==> either x+=+8+ or x+=+-3


So the solution is either+x+=+8++or++x+=+-3++


Hope you found the explanation useful.


Regards.


Praseena.