Solve -2(2x+5)(1-x)<0

$$- 2 ( 2 x + 5 ) ( 1 - x ) < 0$$

Solve for x

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

Steps Using the Quadratic Formula

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

$$\left(-4x-10\right)\left(1-x\right)<0$$

Use the distributive property to multiply -4x-10 by 1-x and combine like terms.

$$6x+4x^{2}-10<0$$

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.

$$6x+4x^{2}-10=0$$

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, 6 for b, and -10 for c in the quadratic formula.

$$x=\frac{-6±\sqrt{6^{2}-4\times 4\left(-10\right)}}{2\times 4}$$

Do the calculations.

$$x=\frac{-6±14}{8}$$

Solve the equation x=\frac{-6±14}{8} when ± is plus and when ± is minus.

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

Rewrite the inequality by using the obtained solutions.

$$4\left(x-1\right)\left(x+\frac{5}{2}\right)<0$$

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

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

This is false for any x.

$$x\in \emptyset$$

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

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

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

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

The final solution is the union of the obtained solutions.

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