Question

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

Answer 1

`\large\boxed{C.\ x=(-2\pm√(34))/(6)}`

Explanation:

`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((2)/(\sqrt6)\right)+\left((2)/(\sqrt6)\right)^2-\left((2)/(\sqrt6)\right)^2=5\qquad\text{add}\ \left((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((2)/(\sqrt6)\right)+\left((2)/(\sqrt6)\right)^2}_((a+b)^2=a^2+2ab+b^2)=\left(x\sqrt6+(2)/(\sqrt6)\right)^2`

`\left(x\sqrt6+(2)/(\sqrt6)\right)^2=5+(4)/(6)\\\\\left(x\sqrt6+(2)/(\sqrt6)\right)^2=5+(2)/(3)\\\\\left(x\sqrt6+(2)/(\sqrt6)\right)^2=(15)/(3)+(2)/(3)\\\\\left(x\sqrt6+(2)/(\sqrt6)\right)^2=(17)/(3)\to x\sqrt6+(2)/(\sqrt6)=\pm\sqrt{(17)/(3)}\qquad\text{multiply both sides by}\ \sqrt6\\\\6x+2=\pm\sqrt{(17)/(3)\cdot6}\\\\6x+2=\pm√((17)(2))\\\\6x+2=\pm√(34)\qquad\text{subtract 2 from both sides}\\\\6x=-2\pm√(34)\qquad\text{divide both sides by 6}\\\\x=(-2\pm√(34))/(6)`