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How to solve 2x^2+3x-2=0

asked Aug 3, 2017 in Algebra by Francis

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=-8/4=-2  and x=2/4 = 1/2

x= -2 and 1/2
answered Aug 3, 2017 by quickwriter (2,510 points)