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Find the number of real solutions for the equation 3x^2_6x+3=0

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3x^2-6x+3=0 ?
whoops +3

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Other answers:

ax^2+bx+c = 0 \[\Delta=b^2-4ac\] if \[\Delta = 0\] then one real root if \[\Delta > 0\] two real roots if \[\Delta < 0\] no real roots
this is called discriminant
so there is none?
or two?
3x^2+6x+3=0 this is the equation ?
no square ?
so what i said isnt relevant
9x + 3 =0 9x = -3 x = -3/9 one solution here
but are you sure there is no x^2 ?!
positive thanks.

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