$2x^2+5x-3 \,=\, 0$ is a given quadratic equation. It is given that we have to solve this quadratic equation by quadratic formula method in this maths problem.

For solving the quadratic equation $2x^2+5x-3 \,=\, 0$, compare this equation with the standard form quadratic equation $ax^2+bx+c \,=\, 0$. It helps us to know the literal coefficients of $x^2$ and $x$ and also the constant.

- $a \,=\, 2$
- $b \,=\, 5$
- $c \,=\, -3$

Now, use the quadratic formula to find the roots or zeros of the quadratic equation by substituting the values of $a$, $b$ and $c$.

$x$ $\,=\,$ $\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$

$\implies$ $x$ $\,=\,$ $\dfrac{-5 \pm \sqrt{5^2-4 \times 2 \times (-3)}}{2 \times 2}$

The substitution of $a$, $b$ and $c$ in quadratic formula gives us an equation and we have to solve it for solving the value of $x$ by simplification.

$\implies$ $x$ $\,=\,$ $\dfrac{-5 \pm \sqrt{25-8 \times (-3)}}{4}$

$\implies$ $x$ $\,=\,$ $\dfrac{-5 \pm \sqrt{25+24}}{4}$

$\implies$ $x$ $\,=\,$ $\dfrac{-5 \pm \sqrt{49}}{4}$

$\implies$ $x$ $\,=\,$ $\dfrac{-5 \pm 7}{4}$

Now, take plus sign in the first case and minus sign in the second case.

$\implies$ $x$ $\,=\,$ $\dfrac{-5+7}{4}$ and $x$ $\,=\,$ $\dfrac{-5-7}{4}$

$\implies$ $x$ $\,=\,$ $\dfrac{2}{4}$ and $x$ $\,=\,$ $\dfrac{-12}{4}$

$\implies$ $x$ $\,=\,$ $\dfrac{\cancel{2}}{\cancel{4}}$ and $x$ $\,=\,$ $\dfrac{-\cancel{12}}{\cancel{4}}$

$\implies$ $x$ $\,=\,$ $\dfrac{1}{2}$ and $x$ $\,=\,$ $-3$

The given quadratic expression $2x^2+5x-3$ is equal to zero when $x \,=\, -3$ and $x \,=\, \dfrac{1}{2}$. Therefore, the solution set for the given quadratic equation is $x \,=\, \Big\{-3, \dfrac{1}{2}\Big\}$

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