Answer:[tex]y=-\frac{3+\sqrt{5}}{2}[/tex] AND [tex]y=-\frac{3-\sqrt{5}}{2}[/tex]Step-by-step explanation:Given: [tex]y^2+3y=-1[/tex]To solve for [tex]y[/tex], we need to get everything on one side of the equal sign and set it to zero. We can do this by adding 1 to both sides. We then get:[tex]y^2+3y+1=0[/tex]We can solve for [tex]y[/tex] by using the quadratic formula:[tex]y=\frac{-b+\sqrt{(b)^2-4(a)(c)}}{2a}[/tex] AND [tex]y=\frac{-b-\sqrt{(b)^2-4(a)(c)}}{2a}[/tex]Let's identify our values:[tex]a: 1\\b: 3\\c: 1[/tex]Plug in the values and simplify.[tex]y=\frac{-3+\sqrt{(3)^2-4(1)(1)}}{2(1)}\\y=\frac{-3+\sqrt{5}}{2}\\-------------------------\\y=\frac{-3-\sqrt{(3)^2-4(1)(1)}}{2(1)}\\\\y=\frac{-3-\sqrt{5}}{2}\\[/tex]Your final answers are:[tex]y=-\frac{3+\sqrt{5}}{2}[/tex] AND [tex]y=-\frac{3-\sqrt{5}}{2}[/tex]