Math Doubts

Solve $2x^2+5x-3 \,=\, 0$ by quadratic formula

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

Compare the given equation with standard form

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.

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

Substitute the values in quadratic formula

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

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

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

Solve the equation by simplifying quadratic formula

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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