Solve {l}{-7x^2-3x}{+4>0}

$$\left. \begin{array} { l } { - 7 x ^ { 2 } - 3 x } \\ { + 4 > 0 } \end{array} \right.$$

Solve for x

$$x\in \left(-1,\frac{4}{7}\right)$$

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in -7x^{2}-3x+4 positive. Since -1 is negative, the inequality direction is changed.

$$7x^{2}+3x-4<0$$

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.

$$7x^{2}+3x-4=0$$

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 7 for a, 3 for b, and -4 for c in the quadratic formula.

$$x=\frac{-3±\sqrt{3^{2}-4\times 7\left(-4\right)}}{2\times 7}$$

Do the calculations.

$$x=\frac{-3±11}{14}$$

Solve the equation x=\frac{-3±11}{14} when ± is plus and when ± is minus.

$$x=\frac{4}{7} x=-1$$

Rewrite the inequality by using the obtained solutions.

$$7\left(x-\frac{4}{7}\right)\left(x+1\right)<0$$

For the product to be negative, x-\frac{4}{7} and x+1 have to be of the opposite signs. Consider the case when x-\frac{4}{7} is positive and x+1 is negative.

$$x-\frac{4}{7}>0 x+1<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x+1 is positive and x-\frac{4}{7} is negative.

$$x+1>0 x-\frac{4}{7}<0$$

The solution satisfying both inequalities is x\in \left(-1,\frac{4}{7}\right).

$$x\in \left(-1,\frac{4}{7}\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(-1,\frac{4}{7}\right)$$