# Solve the equation. 5x^{2}=2x-3

Solve the equation.
$$\displaystyle{5}{x}^{{{2}}}={2}{x}-{3}$$

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Step 1
Given: $$\displaystyle{5}{x}^{{{2}}}={2}{x}-{3}$$
For finding solution of given equation, we do transportation then by using quadratic formula find the solution of given equation
So,
$$\displaystyle{5}{x}^{{{2}}}-{2}{x}+{3}={0}$$...(1)
We know that solution of quadratic equation $$\displaystyle{\left({a}{x}^{{{2}}}+{b}{x}+{c}={0}\right)}$$ is given by
$$\displaystyle{x}={\frac{{-{b}\pm\sqrt{{{b}^{{{2}}}-{4}{a}{c}}}}}{{{2}{a}}}}$$...(2)
Step 2
So, by using equation(2) solution of given equation will be
$$\displaystyle{x}={\frac{{-{\left(-{2}\right)}\pm\sqrt{{{\left(-{2}\right)}^{{{2}}}-{4}{\left({5}\right)}{\left({3}\right)}}}}}{{{2}{\left({5}\right)}}}}$$
$$\displaystyle={\frac{{{4}\pm\sqrt{{{4}-{60}}}}}{{{10}}}}$$
$$\displaystyle={\frac{{{4}\pm\sqrt{{-{56}}}}}{{{10}}}}$$
$$\displaystyle={\frac{{{4}\pm{2}\sqrt{{{14}}}{i}}}{{{10}}}}$$
Hence, solution of given equation is $$\displaystyle{x}={\frac{{{4}\pm{2}\sqrt{{{14}}}{i}}}{{{10}}}}$$.