Solve 3x(2x+1)-5<-10x

$$3 x ( 2 x + 1 ) - 5 < - 10 x$$

Solve for x

$$x\in \left(-\frac{5}{2},\frac{1}{3}\right)$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3x by 2x+1.

$$6x^{2}+3x-5<-10x$$

Add 10x to both sides.

$$6x^{2}+3x-5+10x<0$$

Combine 3x and 10x to get 13x.

$$6x^{2}+13x-5<0$$

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.

$$6x^{2}+13x-5=0$$

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 6 for a, 13 for b, and -5 for c in the quadratic formula.

$$x=\frac{-13±\sqrt{13^{2}-4\times 6\left(-5\right)}}{2\times 6}$$

Do the calculations.

$$x=\frac{-13±17}{12}$$

Solve the equation x=\frac{-13±17}{12} when ± is plus and when ± is minus.

$$x=\frac{1}{3} x=-\frac{5}{2}$$

Rewrite the inequality by using the obtained solutions.

$$6\left(x-\frac{1}{3}\right)\left(x+\frac{5}{2}\right)<0$$

For the product to be negative, x-\frac{1}{3} and x+\frac{5}{2} have to be of the opposite signs. Consider the case when x-\frac{1}{3} is positive and x+\frac{5}{2} is negative.

$$x-\frac{1}{3}>0 x+\frac{5}{2}<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x+\frac{5}{2} is positive and x-\frac{1}{3} is negative.

$$x+\frac{5}{2}>0 x-\frac{1}{3}<0$$

The solution satisfying both inequalities is x\in \left(-\frac{5}{2},\frac{1}{3}\right).

$$x\in \left(-\frac{5}{2},\frac{1}{3}\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(-\frac{5}{2},\frac{1}{3}\right)$$