Solve 4x^2-32>x^2+20x

$$4 x ^ { 2 } - 32 > x ^ { 2 } + 20 x$$

Solve for x

$$x\in \left(-\infty,-\frac{4}{3}\right)\cup \left(8,\infty\right)$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

$$4x^{2}-32-x^{2}>20x$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}-32>20x$$

Subtract 20x from both sides.

$$3x^{2}-32-20x>0$$

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.

$$3x^{2}-32-20x=0$$

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 3 for a, -20 for b, and -32 for c in the quadratic formula.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 3\left(-32\right)}}{2\times 3}$$

Do the calculations.

$$x=\frac{20±28}{6}$$

Solve the equation x=\frac{20±28}{6} when ± is plus and when ± is minus.

$$x=8 x=-\frac{4}{3}$$

Rewrite the inequality by using the obtained solutions.

$$3\left(x-8\right)\left(x+\frac{4}{3}\right)>0$$

For the product to be positive, x-8 and x+\frac{4}{3} have to be both negative or both positive. Consider the case when x-8 and x+\frac{4}{3} are both negative.

$$x-8<0 x+\frac{4}{3}<0$$

The solution satisfying both inequalities is x<-\frac{4}{3}.

$$x<-\frac{4}{3}$$

Consider the case when x-8 and x+\frac{4}{3} are both positive.

$$x+\frac{4}{3}>0 x-8>0$$

The solution satisfying both inequalities is x>8.

$$x>8$$

The final solution is the union of the obtained solutions.

$$x<-\frac{4}{3}\text{; }x>8$$