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# How do you find the zeroes of g(x)=2x^2-5x-3 ?

Jun 7, 2015

$g \left(x\right) = 2 {x}^{2} - 5 x - 3$can be factored as
$\textcolor{w h i t e}{\text{XXXX}}$$g \left(x\right) = \left(2 x + 1\right) \left(x - 3\right)$

The zeroes of $g \left(x\right)$ are the values of $x$ for which $g \left(x\right) = 0$

If $g \left(x\right) = \left(2 x + 1\right) \left(x - 3\right) = 0$ then
either
$\textcolor{w h i t e}{\text{XXXX}}$$2 x + 1 = 0 \Rightarrow x = - \frac{1}{2}$
or
$\textcolor{w h i t e}{\text{XXXX}}$$x - 3 = 0 \Rightarrow x = 3$

So the zeros of $g \left(x\right)$ are $\left(- \frac{1}{2}\right)$ and $3$

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