Solve 4x^2-3>4x

$$4 x ^ { 2 } - 3 > 4 x$$

Solve for x

$$x\in \left(-\infty,-\frac{1}{2}\right)\cup \left(\frac{3}{2},\infty\right)$$

Steps Using the Quadratic Formula

Subtract 4x from both sides.

$$4x^{2}-3-4x>0$$

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.

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

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

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

Do the calculations.

$$x=\frac{4±8}{8}$$

Solve the equation x=\frac{4±8}{8} when ± is plus and when ± is minus.

$$x=\frac{3}{2} x=-\frac{1}{2}$$

Rewrite the inequality by using the obtained solutions.

$$4\left(x-\frac{3}{2}\right)\left(x+\frac{1}{2}\right)>0$$

For the product to be positive, x-\frac{3}{2} and x+\frac{1}{2} have to be both negative or both positive. Consider the case when x-\frac{3}{2} and x+\frac{1}{2} are both negative.

$$x-\frac{3}{2}<0 x+\frac{1}{2}<0$$

The solution satisfying both inequalities is x<-\frac{1}{2}.

$$x<-\frac{1}{2}$$

Consider the case when x-\frac{3}{2} and x+\frac{1}{2} are both positive.

$$x+\frac{1}{2}>0 x-\frac{3}{2}>0$$

The solution satisfying both inequalities is x>\frac{3}{2}.

$$x>\frac{3}{2}$$

The final solution is the union of the obtained solutions.

$$x<-\frac{1}{2}\text{; }x>\frac{3}{2}$$