$2x^2+5x-3 \,=\, 0$ 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 $x \,=\, \Big\{-3, \dfrac{1}{2}\Big\}$

Latest Math Topics

Aug 31, 2024

Aug 07, 2024

Jul 24, 2024

Dec 13, 2023

Latest Math Problems

Oct 22, 2024

Oct 17, 2024

Sep 04, 2024

Jan 30, 2024

Oct 15, 2023

Copyright © 2012 - 2023 Math Doubts, All Rights Reserved