A quadratic equation $2x^2+x-1 = 0$ is given in this problem and its roots (or zeros) should be calculated by using the quadratic formula. So, let’s learn how to find the zeros of the quadratic equation $2x^2+x-1 = 0$.

Let us compare the given quadratic equation $2x^2+x-1 \,=\, 0$ with standard form of the quadratic equation $ax^2+bx+c \,=\,0$.

- $a = 2$
- $b = 1$
- $c = -1$

It is time to substitute the values of them in the quadratic formula.

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

$\implies$ $x \,=\, \dfrac{-1\pm \sqrt{1^2-4 \times 2 \times (-1)}}{2 \times 2}$

Now, simplify the quadratic formula to find the zeros or the given quadratic equation $2x^2+x-1$ $=$ $0$.

$\implies$ $x \,=\, \dfrac{-1\pm \sqrt{1+8}}{4}$

$\implies$ $x \,=\, \dfrac{-1\pm \sqrt{9}}{4}$

$\implies$ $x \,=\, \dfrac{-1\pm 3}{4}$

$\implies$ $x \,=\, \dfrac{-1+3}{4}$ or $x \,=\, \dfrac{-1-3}{4}$

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

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

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

$\,\,\,\therefore\,\,\,\,\,\,$ $x \,=\, \dfrac{1}{2}$ or $x \,=\, -1$

The solution set of the given quadratic equation $2x^2+x-1$ $=$ $0$ is $\bigg\{-1, \dfrac{1}{2}\bigg\}$

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