SOLUTION: i need help solving these quadratic equations using the quadratic formula. 2x^2 - 5x = 3 3x^2 - 2x + 1 = 0 i am trying to solve for x 2(x-5)^2 = 3 and this queston also i

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Question 107047: i need help solving these quadratic equations using the quadratic formula.
2x^2 - 5x = 3
3x^2 - 2x + 1 = 0
i am trying to solve for x
2(x-5)^2 = 3
and this queston also i have been trying to figure these out fo an hour.
x^2 + 4x + 4 = 7
Answer by Annabelle1(69) About Me  (Show Source):
You can put this solution on YOUR website!
lets start with 2x^2 - 5x = 3
we have to bring everything together to one side of the equals sign. we get
2x^2 - 5x -3=0
the coeffiecient of x^2 (2) is our a (a=2)
the coeffiecient of x (-5) is our b (b=-5)
the constant is -3 (c=-3)
the quadratic formula is x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x=%28%285+%2B-+sqrt%2849%29%29%2F4+
x=5+%2B-1.75
x=6.75 or x= 3.25

The next one is 3x^2 - 2x + 1 = 0
here a=3 b= -2 and c=1
x+=+%28-%28-2%29+%2B-+sqrt%28+%28-2%29%5E2-4%2A3%2A%281%29+%29%29%2F%282%2A3%29+
this one has a negative under the square root sign so there is no solution ( the graph never crosses the x axis)

Next....
2(x-5)^2 = 3
expand this out first
we get
2%28x%5E2-10x%2B25%29=3
2x%5E2-20x%2B50-3=0
2x%5E2-20x%2B47=0
a=2 b=-20 c=47
x+=+%28-%28-20%29+%2B-+sqrt%28+%28-20%29%5E2-4%2A2%2A%2847%29+%29%29%2F%282%2A2%29+
you simplify from here
Hope these have helped