As given by the question
There are given that the equation
Now,
From the first equation
[tex]5x^2+2=4x[/tex]
Solve the given equation by using quadratic formula to check the solution is imaginary or real.
So,
From the quadratic formula;
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]
So,
[tex]\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x=\frac{-(-4)\pm\sqrt[]{(-4)^2-4\times5\times2}}{2\times5} \\ x=\frac{4\pm2\sqrt[]{6}i}{2\times5} \\ x=\frac{2}{5}+i\frac{\sqrt[]{6}}{5},\text{ }\frac{2}{5}+i\frac{\sqrt[]{6}}{5} \end{gathered}[/tex]
Hence, the value of the given equation is shown below:
[tex]x=\frac{2}{5}+i\frac{\sqrt[]{6}}{5},\text{ }\frac{2}{5}+i\frac{\sqrt[]{6}}{5}[/tex]
So, there is No real solution.
