Question

What is the solution of (4x+3)^2=18

Answer 1

$(4x+3)^2=18\iff4x+3=\pm\sqrt{18}\\\\4x+3=-\sqrt{9\cdot2}\ \vee\ 4x+3=\sqrt{9\cdot2}\\\\4x+3=-\sqrt9\cdot\sqrt2\ \vee\ 4x+3=\sqrt9\cdot\sqrt2\\\\4x+3=-3\sqrt2\ \vee\ 4x+3=3\sqrt2\ \ \ \ \ \ \ \ |-3\\\\4x=-3-3\sqrt2\ \vee\ 4x=-3+3\sqrt2\ \ \ \ \ \ \ |:4\\\\\boxed{x=\dfrac{-3-3\sqrt2}{4}\ \vee\ x=\dfrac{-3+3\sqrt2}{4}}$