Solve y^2-3y-10>0

$$y ^ { 2 } - 3 y - 10 > 0$$

Solve for y

$$y\in \left(-\infty,-2\right)\cup \left(5,\infty\right)$$

Steps Using the Quadratic Formula

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.

$$y^{2}-3y-10=0$$

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, -3 for b, and -10 for c in the quadratic formula.

$$y=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 1\left(-10\right)}}{2}$$

Do the calculations.

$$y=\frac{3±7}{2}$$

Solve the equation y=\frac{3±7}{2} when ± is plus and when ± is minus.

$$y=5 y=-2$$

Rewrite the inequality by using the obtained solutions.

$$\left(y-5\right)\left(y+2\right)>0$$

For the product to be positive, y-5 and y+2 have to be both negative or both positive. Consider the case when y-5 and y+2 are both negative.

$$y-5<0 y+2<0$$

The solution satisfying both inequalities is y<-2.

$$y<-2$$

Consider the case when y-5 and y+2 are both positive.

$$y+2>0 y-5>0$$

The solution satisfying both inequalities is y>5.

$$y>5$$

The final solution is the union of the obtained solutions.

$$y<-2\text{; }y>5$$