# How do you solve 2x^2 = 3x -2=0?

Jun 25, 2015

I think you meant $2 {x}^{2} + 3 x - 2 = 0$ since $+$ and $=$ are on the same key.

$0 = 2 {x}^{2} + 3 x - 2 = \left(2 x - 1\right) \left(x + 2\right)$

So $x = \frac{1}{2}$ or $x = - 2$

#### Explanation:

Let $f \left(x\right) = 2 {x}^{2} + 3 x - 2$.

If $f \left(x\right)$ has factors with integer coefficients, then they must be of the form $\left(2 x \pm 1\right) \left(x \pm 2\right)$ for some combination of $\pm$ signs.

To see this, first note that $2$ only factors as $\pm 2 \times \pm 1$. The coefficient of the ${x}^{2}$ term is $2$, the constant term is $2$ and the middle term is not divisible by $2$. So one of the factors is $\left(2 x \pm 1\right)$ and the other is $\left(x \pm 2\right)$

Furthermore, the sign of the constant term is $-$, so one of the factors has a $+$ and the other a $-$. That leaves only two possibilities to try:

$\left(2 x + 1\right) \left(x - 2\right) = 2 {x}^{2} - 3 x - 2$

and

$\left(2 x - 1\right) \left(x + 2\right) = 2 {x}^{2} + 3 x - 2$