SOLUTION: this is not in a text book 400(x-1)^2=225 this is what I have 400x^2-800x+175=0 my instructions are solve for x x= -(-800) +/- sqrt[(-800)^2 - 4(400*175)]/2(400) x= [800+/- s

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: this is not in a text book 400(x-1)^2=225 this is what I have 400x^2-800x+175=0 my instructions are solve for x x= -(-800) +/- sqrt[(-800)^2 - 4(400*175)]/2(400) x= [800+/- s      Log On


   

Question 154198: this is not in a text book
400(x-1)^2=225
this is what I have
400x^2-800x+175=0
my instructions are solve for x
x= -(-800) +/- sqrt[(-800)^2 - 4(400*175)]/2(400)
x= [800+/- sqrt(640000-280000)] / 800
x= (800 +/- sqrt 360,000) /800
x= (800 +/- 1.000000001)/800
is it right up to this point? where do we go from here?
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

400%28x-1%29%5E2=225 Start with the given equation.


%28x-1%29%5E2=%28225%29%2F%28400%29 Divide both sides by 400.


%28x-1%29%5E2=9%2F16 Reduce.


x-1=0%2B-sqrt%289%2F16%29 Take the square root of both sides.


x-1=sqrt%289%2F16%29 or x-1=-sqrt%289%2F16%29 Break up the "plus/minus" to form two equations.


x-1=3%2F4 or x-1=-3%2F4 Take the square root of 9%2F16 to get 3%2F4.


x=1%2B3%2F4 or x=1-3%2F4 Add 1 to both sides.


x=7%2F4 or x=1%2F4 Combine like terms.


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Answer:


So the solutions are x=7%2F4 or x=1%2F4.