Question

Please help me with #'s 5&6

Answer 1

$5.\\(x+3)^2+7=-2\qquad|-7\\\\(x+3)^2=-9 < 0\\\\\text{NO REAL SOLUTION}\\\\(x+3)^2=-9\to x+3=\pm\sqrt{-9}\\\\x+3=-3i\ \vee\ x+3=3i\qquad|-3\\\\\boxed{x=-3-3i\ \vee\ x=-3+3i}$

$6.\\4(x-10)^2=25\qquad|:4\\\\(x-10)^2=\dfrac{25}{4}\to x-10=\pm\sqrt{\dfrac{25}{4}}\\\\x-10=-\dfrac{\sqrt{25}}{\sqrt4}\ \vee\ x-10=\dfrac{\sqrt{25}}{\sqrt4}\\\\x-10=-\dfrac{5}{2}\ \vee\ x-10=\dfrac{5}{2}\\\\x-10=-2.5\ \vee\ x-10=2.5\qquad|+10\\\\\boxed{x=7.5\ \vee\ x=12.5}$