What is the answer to this question and how do you figure it out

Answer:[tex]\large\boxed{C.\ x=\dfrac{-2\pm\sqrt{34}}{6}}[/tex]Step-by-step explanation:[tex]6x^2+4x-5=0\qquad\text{add 5 to both sides}\\\\(x\sqrt6)^2+2(2x)=5\\\\(x\sqrt6)^2+2(x\sqrt6)\left(\dfrac{2}{\sqrt6}\right)+\left(\dfrac{2}{\sqrt6}\right)^2-\left(\dfrac{2}{\sqrt6}\right)^2=5\qquad\text{add}\ \left(\dfrac{2}{\sqrt6}\right)^2\ \text{to both sides}\\\\\text{use}\ (a+b)^2=a^2+2ab+b^2\\\downarrow\\\underbrace{(x\sqrt6)^2+2(x\sqrt6)\left(\dfrac{2}{\sqrt6}\right)+\left(\dfrac{2}{\sqrt6}\right)^2}_{(a+b)^2=a^2+2ab+b^2}=\left(x\sqrt6+\dfrac{2}{\sqrt6}\right)^2[/tex][tex]\left(x\sqrt6+\dfrac{2}{\sqrt6}\right)^2=5+\dfrac{4}{6}\\\\\left(x\sqrt6+\dfrac{2}{\sqrt6}\right)^2=5+\dfrac{2}{3}\\\\\left(x\sqrt6+\dfrac{2}{\sqrt6}\right)^2=\dfrac{15}{3}+\dfrac{2}{3}\\\\\left(x\sqrt6+\dfrac{2}{\sqrt6}\right)^2=\dfrac{17}{3}\to x\sqrt6+\dfrac{2}{\sqrt6}=\pm\sqrt{\dfrac{17}{3}}\qquad\text{multiply both sides by}\ \sqrt6\\\\6x+2=\pm\sqrt{\dfrac{17}{3}\cdot6}\\\\6x+2=\pm\sqrt{(17)(2)}\\\\6x+2=\pm\sqrt{34}\qquad\text{subtract 2 from both sides}\\\\6x=-2\pm\sqrt{34}\qquad\text{divide both sides by 6}\\\\x=\dfrac{-2\pm\sqrt{34}}{6}[/tex]