SOLUTION: How would I use the quadratic formula to solve for x^2-9x+18=0?

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Question 947605: How would I use the quadratic formula to solve for x^2-9x+18=0?
Found 3 solutions by MathLover1, rfer, josmiceli:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!


x%5E2-9x%2B18=0

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+ to use the quadratic formula to solve for, note that a=1,b=-9, and c=18; so, plug it in
x+=+%28-%28-9%29+%2B-+sqrt%28+%28-9%29%5E2-4%2A1%2A18+%29%29%2F%282%2A1%29+

x+=+%289+%2B-+sqrt%28+81-72+%29%29%2F2+

x+=+%289+%2B-+sqrt%28+9+%29%29%2F2+

x+=+%289+%2B-+3%29%2F2+
solutions:
x+=+%289+%2B+3%29%2F2+
x+=+12%2F2+
x+=+6+
or
x+=+%289+-+3%29%2F2+
x+=+6%2F2+
x+=+3+

+graph%28+600%2C+600%2C+-10%2C+10%2C+-10%2C+20%2C+x%5E2-9x%2B18%29+


Answer by rfer(16322) About Me  (Show Source):
You can put this solution on YOUR website!
a=1
b=-9
c=18
---------------
dis=
sqrt(-9^2-4*1*18)
=3
----------------
(--9+-3)/(2*1)
9-3/2=3
9+3/2=6
(0,3)
(0,6)

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+x%5E2++-+9x+%2B+18+=+0+
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
+a+=+1+
+b+=+-9+
+c+=+18+
------------
x+=+%28-%28-9%29+%2B-+sqrt%28+%28-9%29%5E2+-+4%2A1%2A18+%29%29%2F%282%2A1%29+
x+=+%28+9+%2B-+sqrt%28+81+-+72+%29%29%2F2+
+x+=+%28+9+%2B-+sqrt%28+9+%29+%29+%2F+2+
+x+=+%28+9+%2B+3+%29+%2F+2+
+x+=+12%2F2+
+x+=+6+
And, using the negative square root,
+x+=+%28+9+-+3+%29+%2F+2+
+x+=+6%2F2+
+x+=+3+
---------------
check:
The factors are:
+%28+x+-+6+%29%2A%28+x+-+3+%29+=+x%5E2+-+9x+%2B+18+
OK