Solve x^2+(3-2x-8/7)x+4-8/7<0

$$x ^ { 2 } + ( 3 - 2 x - \frac { 8 } { 7 } ) x + 4 - \frac { 8 } { 7 } < 0$$

Solve for x

$$x\in \left(-\infty,-1\right)\cup \left(\frac{20}{7},\infty\right)$$

Steps Using the Quadratic Formula

Subtract \frac{8}{7} from 3 to get \frac{13}{7}.

$$x^{2}+\left(\frac{13}{7}-2x\right)x+4-\frac{8}{7}<0$$

Use the distributive property to multiply \frac{13}{7}-2x by x.

$$x^{2}+\frac{13}{7}x-2x^{2}+4-\frac{8}{7}<0$$

Combine x^{2} and -2x^{2} to get -x^{2}.

$$-x^{2}+\frac{13}{7}x+4-\frac{8}{7}<0$$

Subtract \frac{8}{7} from 4 to get \frac{20}{7}.

$$-x^{2}+\frac{13}{7}x+\frac{20}{7}<0$$

Multiply the inequality by -1 to make the coefficient of the highest power in -x^{2}+\frac{13}{7}x+\frac{20}{7} positive. Since -1 is negative, the inequality direction is changed.

$$x^{2}-\frac{13}{7}x-\frac{20}{7}>0$$

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.

$$x^{2}-\frac{13}{7}x-\frac{20}{7}=0$$

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 1 for a, -\frac{13}{7} for b, and -\frac{20}{7} for c in the quadratic formula.

$$x=\frac{-\left(-\frac{13}{7}\right)±\sqrt{\left(-\frac{13}{7}\right)^{2}-4\times 1\left(-\frac{20}{7}\right)}}{2}$$

Do the calculations.

$$x=\frac{\frac{13}{7}±\frac{27}{7}}{2}$$

Solve the equation x=\frac{\frac{13}{7}±\frac{27}{7}}{2} when ± is plus and when ± is minus.

$$x=\frac{20}{7} x=-1$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-\frac{20}{7}\right)\left(x+1\right)>0$$

For the product to be positive, x-\frac{20}{7} and x+1 have to be both negative or both positive. Consider the case when x-\frac{20}{7} and x+1 are both negative.

$$x-\frac{20}{7}<0 x+1<0$$

The solution satisfying both inequalities is x<-1.

$$x<-1$$

Consider the case when x-\frac{20}{7} and x+1 are both positive.

$$x+1>0 x-\frac{20}{7}>0$$

The solution satisfying both inequalities is x>\frac{20}{7}.

$$x>\frac{20}{7}$$

The final solution is the union of the obtained solutions.

$$x<-1\text{; }x>\frac{20}{7}$$