Solve a^2+3a-10<0

$$a ^ { 2 } + 3 a - 10 < 0$$

Solve for a

$$a\in \left(-5,2\right)$$

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.

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

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.

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

Do the calculations.

$$a=\frac{-3±7}{2}$$

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

$$a=2 a=-5$$

Rewrite the inequality by using the obtained solutions.

$$\left(a-2\right)\left(a+5\right)<0$$

For the product to be negative, a-2 and a+5 have to be of the opposite signs. Consider the case when a-2 is positive and a+5 is negative.

$$a-2>0 a+5<0$$

This is false for any a.

$$a\in \emptyset$$

Consider the case when a+5 is positive and a-2 is negative.

$$a+5>0 a-2<0$$

The solution satisfying both inequalities is a\in \left(-5,2\right).

$$a\in \left(-5,2\right)$$

The final solution is the union of the obtained solutions.

$$a\in \left(-5,2\right)$$