Expert: Ahmed Salami - 8/11/2004

Question
    Hi there Ahmed.  How are you doing today?  I hope all is well with you.  I'm having difficulty with these problems.  Can you help me?  It says solve using the quadratic formula.  I'm lost.

1.  3x2  -  6x  =  24

2.  x2  +  10x  -  3  =  0

Answer
Hi Damon,
I'm terribly sorry about the delay.
For a quadratic equation, ax^2 + bx + c = 0, the roots are
(1/2a)[-b +or- sqrt(b^2 - 4ac)]
For the first one, 3x^2 - 6x - 24 = 0
a = 3, b = -6, c = -24
the roots are
(1/6)[6 +or- sqrt(36 - -288)]
= (1/6)[6 +or- sqrt(324)]
= (1/6)(6 +or- 18)
= (1/6)(24) or (1/6)(-12)
= 4 or -2

for the other one,  x^2 + 10x - 3 = 0
a = 1, b = 10, c = -3
the roots are
(1/2)[-10 +or- sqrt(100 - -12)]
= (1/2)[-10 +or- sqrt(112)]
= (1/2)[-10 +or- 10.58]
= (1/2)(0.58)or (1/2)(-20.58)
= 0.29 or -10.29

I hope that does it.
Good luck.
Regards.