Solve 3+2y-y^2<0

$$3 + 2 y - y ^ { 2 } < 0$$

Solve for y

$$y\in \left(-\infty,-1\right)\cup \left(3,\infty\right)$$

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in 3+2y-y^{2} positive. Since -1 is negative, the inequality direction is changed.

$$-3-2y+y^{2}>0$$

To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$-3-2y+y^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, -2 for b, and -3 for c in the quadratic formula.

$$y=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 1\left(-3\right)}}{2}$$

Do the calculations.

$$y=\frac{2±4}{2}$$

Solve the equation y=\frac{2±4}{2} when ± is plus and when ± is minus.

$$y=3 y=-1$$

Rewrite the inequality by using the obtained solutions.

$$\left(y-3\right)\left(y+1\right)>0$$

For the product to be positive, y-3 and y+1 have to be both negative or both positive. Consider the case when y-3 and y+1 are both negative.

$$y-3<0 y+1<0$$

The solution satisfying both inequalities is y<-1.

$$y<-1$$

Consider the case when y-3 and y+1 are both positive.

$$y+1>0 y-3>0$$

The solution satisfying both inequalities is y>3.

$$y>3$$

The final solution is the union of the obtained solutions.

$$y<-1\text{; }y>3$$