Question

(4x)^2 - 3x (3x+9) = 13-7x.
Give answer in simplified radical form.

Answer 1

$$(4x)^2-3x(3x+9)=13-7x\\16x^2-9x^2-27x=13-7x\\16x^2-9x^2-27x+7x-13=0\\7x^2-20x-13=0\\x=\frac{-(-20)\pm\sqrt{(-20)^2-4(7)(-13)}}{2(7)}=\frac{20\pm\sqrt{400+364}}{14}=\frac{20\pm\sqrt{764}}{14}$$
$$\frac{20\pm\sqrt{764}}{14}=\frac{10\pm\sqrt{191}}7$$