Question 300155
Quadratic Equation Form: y = ax^2 + bc + c
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Quadratic Formula: x = [-b +- sqrt(b^2-4ac)]/(2a)
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i dont understand some of the quadradic equation steps, like:
* when to use the two different equations.
I think you mean "two different solutions".
Every quadratic has two solutions.
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*when do i know that there are no solutions
I think you mean "no Real Number solutions".
If b^2-4ac is negative the equation has no Real Number solutions.
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*if i got a fraction for X do i turn it into a decimanl?
That is your choice.  You may change it or you may leave it as a fraction.
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*if i dont have to turn it into decimal is there a chance that i can get a real big number? how can i work that out?
Yes, solutions may be big, or they may be small.
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*when i use the long equation(the one with the square root on it) im looking for the vertex in the parabola or what.. i dont understand?
You are looking for the zeroes or roots of the equation.
You are looking for the values of "x" which make ax^2+bx+c = 0
You are looking for the x-intercepts of the graph of the equation.
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Cheers,
Stan H.
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