Solve 4x^2+x-3<0

$$4 x ^ { 2 } + x - 3 < 0$$

Solve for x

$$x\in \left(-1,\frac{3}{4}\right)$$

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.

$$4x^{2}+x-3=0$$

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 4 for a, 1 for b, and -3 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{-1±7}{8}$$

Solve the equation x=\frac{-1±7}{8} when ± is plus and when ± is minus.

$$x=\frac{3}{4} x=-1$$

Rewrite the inequality by using the obtained solutions.

$$4\left(x-\frac{3}{4}\right)\left(x+1\right)<0$$

For the product to be negative, x-\frac{3}{4} and x+1 have to be of the opposite signs. Consider the case when x-\frac{3}{4} is positive and x+1 is negative.

$$x-\frac{3}{4}>0 x+1<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x+1 is positive and x-\frac{3}{4} is negative.

$$x+1>0 x-\frac{3}{4}<0$$

The solution satisfying both inequalities is x\in \left(-1,\frac{3}{4}\right).

$$x\in \left(-1,\frac{3}{4}\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(-1,\frac{3}{4}\right)$$