Answer
$\frac{3}{4}$.
Work Step by Step
$\sqrt{4-5x}=2x-1$ …… (1)
Take the square of both sides.
$4-5x={{\left( 2x-1 \right)}^{2}}$
Apply the formula for the square of difference of any two numbers and simplify the equation.
$\begin{align}
& 4-5x=4{{x}^{2}}-4x+1 \\
& 4{{x}^{2}}+x-3=0 \\
& 4x\left( x+1 \right)-3\left( x+1 \right)=0 \\
& \left( 4x-3 \right)\left( x+1 \right)=0
\end{align}$
The solution is,
$\begin{align}
& 4x-3=0 \\
& x=\frac{3}{4}
\end{align}$
Or,
$\begin{align}
& x+1=0 \\
& x=-1
\end{align}$
Check:
Substitute $x=\frac{3}{4}$ in equation (1).
$\begin{matrix}
\sqrt{4-5\left( \frac{3}{4} \right)}\overset{?}{\mathop{=}}\,2\left( \frac{3}{4} \right)-1 \\
\sqrt{\frac{1}{4}}\overset{?}{\mathop{=}}\,\frac{3}{2}-1 \\
\frac{1}{2}=\frac{1}{2} \\
\end{matrix}$
It is true.
Substitute $x=-1$ in equation (1).
$\begin{matrix}
\sqrt{4-5\left( -1 \right)}\overset{?}{\mathop{=}}\,2\left( -1 \right)-1 \\
\sqrt{9}\overset{?}{\mathop{=}}\,-3 \\
3=-3 \\
\end{matrix}$
It is false.
