Solve 2x(x+3)-2(3x+5)+x>0

$$2 x ( x + 3 ) - 2 ( 3 x + 5 ) + x > 0$$

Solve for x

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

Steps Using the Quadratic Formula

Use the distributive property to multiply 2x by x+3.

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

Use the distributive property to multiply -2 by 3x+5.

$$2x^{2}+6x-6x-10+x>0$$

Combine 6x and -6x to get 0.

$$2x^{2}-10+x>0$$

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.

$$2x^{2}-10+x=0$$

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 2 for a, 1 for b, and -10 for c in the quadratic formula.

$$x=\frac{-1±\sqrt{1^{2}-4\times 2\left(-10\right)}}{2\times 2}$$

Do the calculations.

$$x=\frac{-1±9}{4}$$

Solve the equation x=\frac{-1±9}{4} when ± is plus and when ± is minus.

$$x=2 x=-\frac{5}{2}$$

Rewrite the inequality by using the obtained solutions.

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

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

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

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

$$x<-\frac{5}{2}$$

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

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

The solution satisfying both inequalities is x>2.

$$x>2$$

The final solution is the union of the obtained solutions.

$$x<-\frac{5}{2}\text{; }x>2$$