SOLUTION: Could you please help me with this problem? Solve using the quadratic Formaula. Give exact solutions 4x^2= -20x-12 Thanks you!!!!!!!!

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Question 208682:
Could you please help me with this problem?
Solve using the quadratic Formaula. Give exact solutions
4x^2= -20x-12
Thanks you!!!!!!!!
Answer by algebrapro18(249) About Me  (Show Source):
You can put this solution on YOUR website!
First off we need to get this equation into standard form which is ax^2+bx+c=0. To do that we need to add 20x and 12 to both sides.

4x^2+20x+12=0. Now we can plug into the quadratic formula.
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

where a = 4, b = 20, and c = 12. Plugging in you get:
x+=+%28-20+%2B-+sqrt%28+20%5E2-4%2A4%2A12+%29%29%2F%282%2A4%29+

Now we simplify:
x+=+%28-20+%2B-+sqrt%28+20%5E2-4%2A4%2A12+%29%29%2F%282%2A4%29+
x+=+%28-20+%2B-+sqrt%28+400-192+%29%29%2F%288%29+
x+=+%28-20+%2B-+sqrt%28+208+%29%29%2F%288%29+

208 is divisible by 16 so you can take divide by 16 and then pull a 4 out because the square root of 16 is 4.
So the expression above simplifies down to

x+=+%28-20+%2B-+4%2Asqrt%28+13+%29%29%2F%288%29+

Since there is a factor of 4 present in all 3 whole numbers we can divide everything by 4. Doing so will result in:

x+=+%28-5+%2B-+sqrt%2813%29%29%2F%282%29+

And that is your final exact answer.