what is the solution set of $x^{2}-10 = 30x$?\n{-220, 250}\n{-250, 220}\n{15 - \\sqrt{235}, 15 +…

Explanation:

Step1: Rearrange to standard quadratic form

$x^{2}-30x - 10=0$

Step2: Identify coefficients for quadratic formula

For $ax^{2}+bx + c = 0$, here $a = 1$, $b=-30$, $c=-10$

Step3: Apply quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

$x=\frac{30\pm\sqrt{(-30)^{2}-4\times1\times(-10)}}{2\times1}=\frac{30\pm\sqrt{900 + 40}}{2}=\frac{30\pm\sqrt{940}}{2}=\frac{30\pm2\sqrt{235}}{2}=15\pm\sqrt{235}$

Answer:

C. ${15-\sqrt{235},15 + \sqrt{235}}$