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Solve by factoring2x^2+3x=2

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The given equation is,


2x^2+3x=2

Converting to the standard quadratic form, ax² + bx + c = 0, we have,


2x^2+3x-2=0

We have the solution for the quadratic equation as,


x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}

Here,


\begin{gathered} a=2 \\ b=3 \\ c=-2 \end{gathered}

Substituting, we get,


\begin{gathered} x=\frac{-3\pm\sqrt[]{3^2-(4)(2)(-2)}}{2*2} \\ =\frac{-3\pm\sqrt[]{9+16}}{4} \\ =(-3\pm5)/(4) \\ =(-3+5)/(4);(-3-5)/(4) \\ =(1)/(2);-2 \end{gathered}

User Ncoronges
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