Question

(3x+2)∧2=9 how do i solve this in the square root property

Answer 1

`(3x+2)^2=9 \\\\9x^2+12x+4-9=0\\\\9x^2+12x-5=0\\\\a=9,\ \ b=12, \ \ c=-5`


`x_(1)=(-b-√(b^2-4ac))/(2a)=(-12-√(12^2-4 \cdot9 \cdot (-5)))/(2 \cdot 9)=(-12-√(144+180))/(18)=\\\\=(-12-√(324))/(18)=(-12-18)/(18)=(-30)/(18)=-(5)/(3)\\\\x_(2)=(-b+√(b^2-4ac))/(2a)=(-12+√(12^2-4 \cdot9 \cdot (-5)))/(2 \cdot 9)=(-12+18)/(18)=(6)/(18)=(1)/(3) \\\\Answer: \ x=-(5)/(3)\ \ and \ \ x=(1)/(3)`