Solve x^2+3x-10>=0

$${ x }^{ 2 } +3x-10 > =0$$

Solve for x

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

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.

$$x^{2}+3x-10=0$$

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.

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

Do the calculations.

$$x=\frac{-3±7}{2}$$

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

$$x=2 x=-5$$

Rewrite the inequality by using the obtained solutions.

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

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

$$x-2<0 x+5<0$$

The solution satisfying both inequalities is x<-5.

$$x<-5$$

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

$$x+5>0 x-2>0$$

The solution satisfying both inequalities is x>2.

$$x>2$$

The final solution is the union of the obtained solutions.

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