## Algebra and Trigonometry 10th Edition

Solve the equation. Check if any of the solutions are extraneous. 1 does not satisfy the equation since the square root of a positive number is positive. $\sqrt {x+10}=x-2$ $x+10=x^2-4x+4$ $x^2-5x-6=0$ (x-6)(x+1)=0 x=6,-1 (-1 is extraneous) x=6