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How to solve 2x^2+3x-2=0
asked
Aug 3, 2017
in
Algebra
by
Francis
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1 Answer
you can always try to solve it through the quadratic equation:
x = (-b +/- sqrt(b^2 -4ac))/2a
so you've got ax^2 + bx + c
a=2, b=3, c=-2
x=(-3+/- sqrt(3^2-(4(2)(-2)))/2(2)
x=(-3+/- sqrt(9+16))/4
x=(-3+/-5)/4
x=-8/4=-2 and x=2/4 = 1/2
x= -2 and 1/2
answered
Aug 3, 2017
by
quickwriter
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2,510
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