Solve for \( u \), where \( u \) is a real number. \[ 3 u=\sqrt{10 u^{2}-3 u-10} \] If there is more than one solution, separate them with commas. If there is no solution, click on "No solution". \( u=\square \)

To solve the equation \( 3u = \sqrt{10u^2 - 3u - 10} \), we can start by squaring both sides to eliminate the square root: \[ (3u)^2 = 10u^2 - 3u - 10 \] This simplifies to: \[ 9u^2 = 10u^2 - 3u - 10 \] Next, move all terms to one side of the equation: \[ 9u^2 - 10u^2 + 3u + 10 = 0 \] This leads to: \[ -u^2 + 3u + 10 = 0 \] Multiplying the entire equation by -1 to make the leading coefficient positive gives: \[ u^2 - 3u - 10 = 0 \] Now, we can apply the quadratic formula where \( a = 1 \), \( b = -3 \), and \( c = -10 \): \[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-10)}}{2 \cdot 1} \] Calculating the discriminant: \[ (-3)^2 - 4 \cdot 1 \cdot (-10) = 9 + 40 = 49 \] Now substituting back into the quadratic formula: \[ u = \frac{3 \pm \sqrt{49}}{2} = \frac{3 \pm 7}{2} \] This results in two solutions: 1. \( u = \frac{10}{2} = 5 \) 2. \( u = \frac{-4}{2} = -2 \) Now, we check both solutions in the original equation to ensure they do not produce extraneous results. For \( u = 5 \): \[ 3(5) = \sqrt{10(5^2) - 3(5) - 10} \implies 15 = \sqrt{250 - 15 - 10} \implies 15 = \sqrt{225} \implies 15 = 15 \quad \text{(valid)} \] For \( u = -2 \): \[ 3(-2) = \sqrt{10(-2)^2 - 3(-2) - 10} \implies -6 = \sqrt{40 + 6 - 10} \implies -6 = \sqrt{36} \implies -6 = 6 \quad \text{(not valid)} \] Thus, the only valid solution is: \[ u = 5 \] Final answer: \( u = 5 \)