Solve 8*-x^2-3x+4>0

$$8 \cdot - x ^ { 2 } - 3 x + 4 > 0$$

Solve for x

$$x\in \left(\frac{-\sqrt{137}-3}{16},\frac{\sqrt{137}-3}{16}\right)$$

Steps Using the Quadratic Formula

Multiply 8 and -1 to get -8.

$$-8x^{2}-3x+4>0$$

Multiply the inequality by -1 to make the coefficient of the highest power in -8x^{2}-3x+4 positive. Since -1 is negative, the inequality direction is changed.

$$8x^{2}+3x-4<0$$

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.

$$8x^{2}+3x-4=0$$

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

$$x=\frac{-3±\sqrt{3^{2}-4\times 8\left(-4\right)}}{2\times 8}$$

Do the calculations.

$$x=\frac{-3±\sqrt{137}}{16}$$

Solve the equation x=\frac{-3±\sqrt{137}}{16} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{137}-3}{16} x=\frac{-\sqrt{137}-3}{16}$$

Rewrite the inequality by using the obtained solutions.

$$8\left(x-\frac{\sqrt{137}-3}{16}\right)\left(x-\frac{-\sqrt{137}-3}{16}\right)<0$$

For the product to be negative, x-\frac{\sqrt{137}-3}{16} and x-\frac{-\sqrt{137}-3}{16} have to be of the opposite signs. Consider the case when x-\frac{\sqrt{137}-3}{16} is positive and x-\frac{-\sqrt{137}-3}{16} is negative.

$$x-\frac{\sqrt{137}-3}{16}>0 x-\frac{-\sqrt{137}-3}{16}<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\frac{-\sqrt{137}-3}{16} is positive and x-\frac{\sqrt{137}-3}{16} is negative.

$$x-\frac{-\sqrt{137}-3}{16}>0 x-\frac{\sqrt{137}-3}{16}<0$$

The solution satisfying both inequalities is x\in \left(\frac{-\sqrt{137}-3}{16},\frac{\sqrt{137}-3}{16}\right).

$$x\in \left(\frac{-\sqrt{137}-3}{16},\frac{\sqrt{137}-3}{16}\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(\frac{-\sqrt{137}-3}{16},\frac{\sqrt{137}-3}{16}\right)$$