Question

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

Answer 1

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

$x_{1}=\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{-12-\sqrt{12^2-4 \cdot9 \cdot (-5)}}{2 \cdot 9}=\frac{-12-\sqrt{144+180}}{18}=\\\\=\frac{-12-\sqrt{324}}{18}=\frac{-12-18}{18}=\frac{-30}{18}=-\frac{5}{3}\\\\x_{2}=\frac{-b+\sqrt{b^2-4ac}}{2a}=\frac{-12+\sqrt{12^2-4 \cdot9 \cdot (-5)}}{2 \cdot 9}=\frac{-12+18}{18}=\frac{6}{18}=\frac{1}{3} \\\\Answer: \ x=-\frac{5}{3}\ \ and \ \ x=\frac{1}{3}$