Solve the following equation for \( z \) \[ 4 z-1=\sqrt{25 z^{2}-14 z+2} \]

To solve the equation \( 4z - 1 = \sqrt{25z^2 - 14z + 2} \), we start by squaring both sides to eliminate the square root. First, rewrite the equation: \[ (4z - 1)^2 = 25z^2 - 14z + 2 \] Next, expand the left side: \[ (4z - 1)(4z - 1) = 16z^2 - 8z + 1 \] So the full equation now looks like: \[ 16z^2 - 8z + 1 = 25z^2 - 14z + 2 \] Rearranging gives us: \[ 0 = 25z^2 - 14z + 2 - 16z^2 + 8z - 1 \] \[ 0 = 9z^2 - 6z + 1 \] Now, we can simplify it a bit: \[ 9z^2 - 6z + 1 = 0 \] This quadratic can be solved using the quadratic formula: \[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Where \( a = 9 \), \( b = -6 \), and \( c = 1 \): \[ z = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 9 \cdot 1}}{2 \cdot 9} \] \[ z = \frac{6 \pm \sqrt{36 - 36}}{18} \] \[ z = \frac{6 \pm 0}{18} \] \[ z = \frac{6}{18} = \frac{1}{3} \] Now we should check if \( z = \frac{1}{3} \) is a solution in the original equation: Substituting \( z = \frac{1}{3} \) into the left side gives: \[ 4 \left( \frac{1}{3} \right) - 1 = \frac{4}{3} - 1 = \frac{4}{3} - \frac{3}{3} = \frac{1}{3} \] Now the right-hand side: \[ \sqrt{25 \left( \frac{1}{3} \right)^2 - 14 \cdot \frac{1}{3} + 2} \] Calculating: \[ \sqrt{25 \cdot \frac{1}{9} - \frac{14}{3} + 2} = \sqrt{\frac{25}{9} - \frac{42}{9} + \frac{18}{9}} = \sqrt{\frac{25 - 42 + 18}{9}} = \sqrt{\frac{1}{9}} = \frac{1}{3} \] Both sides equal \( \frac{1}{3} \). Therefore, the solution is: \[ \boxed{\frac{1}{3}} \]