In simplest radical form, what are the solutions to the quadratic equation 6=x2−10⁢x? Quadratic formula: x is equal to the fraction with numerator negative b plus or minus square root of b squared minus 4 A c end root and denominator 2 A

Rewrite the equation as x² − 10x − 6 = 0, so a = 1, b = −10, and c = −6. The quadratic formula gives x = (10 ± √(100 + 24))/2 = (10 ± √124)/2. Since √124 = 2√31, the solutions simplify to x = 5 ± √31, which is choice C.

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