Solve 5a^2+2a-3<0

$$5 a ^ { 2 } + 2 a - 3 < 0$$

Solve for a

$$a\in \left(-1,\frac{3}{5}\right)$$

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.

$$5a^{2}+2a-3=0$$

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 5 for a, 2 for b, and -3 for c in the quadratic formula.

$$a=\frac{-2±\sqrt{2^{2}-4\times 5\left(-3\right)}}{2\times 5}$$

Do the calculations.

$$a=\frac{-2±8}{10}$$

Solve the equation a=\frac{-2±8}{10} when ± is plus and when ± is minus.

$$a=\frac{3}{5} a=-1$$

Rewrite the inequality by using the obtained solutions.

$$5\left(a-\frac{3}{5}\right)\left(a+1\right)<0$$

For the product to be negative, a-\frac{3}{5} and a+1 have to be of the opposite signs. Consider the case when a-\frac{3}{5} is positive and a+1 is negative.

$$a-\frac{3}{5}>0 a+1<0$$

This is false for any a.

$$a\in \emptyset$$

Consider the case when a+1 is positive and a-\frac{3}{5} is negative.

$$a+1>0 a-\frac{3}{5}<0$$

The solution satisfying both inequalities is a\in \left(-1,\frac{3}{5}\right).

$$a\in \left(-1,\frac{3}{5}\right)$$

The final solution is the union of the obtained solutions.

$$a\in \left(-1,\frac{3}{5}\right)$$