Solve 5x(x-1)-3>2x-x

$$5 x ( x - 1 ) - 3 > 2 x - x$$

Solve for x

$$x\in \left(-\infty,\frac{3-2\sqrt{6}}{5}\right)\cup \left(\frac{2\sqrt{6}+3}{5},\infty\right)$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 5x by x-1.

$$5x^{2}-5x-3>2x-x$$

Combine 2x and -x to get x.

$$5x^{2}-5x-3>x$$

Subtract x from both sides.

$$5x^{2}-5x-3-x>0$$

Combine -5x and -x to get -6x.

$$5x^{2}-6x-3>0$$

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.

$$5x^{2}-6x-3=0$$

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, -6 for b, and -3 for c in the quadratic formula.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 5\left(-3\right)}}{2\times 5}$$

Do the calculations.

$$x=\frac{6±4\sqrt{6}}{10}$$

Solve the equation x=\frac{6±4\sqrt{6}}{10} when ± is plus and when ± is minus.

$$x=\frac{2\sqrt{6}+3}{5} x=\frac{3-2\sqrt{6}}{5}$$

Rewrite the inequality by using the obtained solutions.

$$5\left(x-\frac{2\sqrt{6}+3}{5}\right)\left(x-\frac{3-2\sqrt{6}}{5}\right)>0$$

For the product to be positive, x-\frac{2\sqrt{6}+3}{5} and x-\frac{3-2\sqrt{6}}{5} have to be both negative or both positive. Consider the case when x-\frac{2\sqrt{6}+3}{5} and x-\frac{3-2\sqrt{6}}{5} are both negative.

$$x-\frac{2\sqrt{6}+3}{5}<0 x-\frac{3-2\sqrt{6}}{5}<0$$

The solution satisfying both inequalities is x<\frac{3-2\sqrt{6}}{5}.

$$x<\frac{3-2\sqrt{6}}{5}$$

Consider the case when x-\frac{2\sqrt{6}+3}{5} and x-\frac{3-2\sqrt{6}}{5} are both positive.

$$x-\frac{3-2\sqrt{6}}{5}>0 x-\frac{2\sqrt{6}+3}{5}>0$$

The solution satisfying both inequalities is x>\frac{2\sqrt{6}+3}{5}.

$$x>\frac{2\sqrt{6}+3}{5}$$

The final solution is the union of the obtained solutions.

$$x<\frac{3-2\sqrt{6}}{5}\text{; }x>\frac{2\sqrt{6}+3}{5}$$