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4x+8-x^{2}>-6x
Subtract x^{2} from both sides.
4x+8-x^{2}+6x>0
Add 6x to both sides.
10x+8-x^{2}>0
Combine 4x and 6x to get 10x.
-10x-8+x^{2}<0
Multiply the inequality by -1 to make the coefficient of the highest power in 10x+8-x^{2} positive. Since -1 is negative, the inequality direction is changed.
-10x-8+x^{2}=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=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 1\left(-8\right)}}{2}
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, -10 for b, and -8 for c in the quadratic formula.
x=\frac{10±2\sqrt{33}}{2}
Do the calculations.
x=\sqrt{33}+5 x=5-\sqrt{33}
Solve the equation x=\frac{10±2\sqrt{33}}{2} when ± is plus and when ± is minus.
\left(x-\left(\sqrt{33}+5\right)\right)\left(x-\left(5-\sqrt{33}\right)\right)<0
Rewrite the inequality by using the obtained solutions.
x-\left(\sqrt{33}+5\right)>0 x-\left(5-\sqrt{33}\right)<0
For the product to be negative, x-\left(\sqrt{33}+5\right) and x-\left(5-\sqrt{33}\right) have to be of the opposite signs. Consider the case when x-\left(\sqrt{33}+5\right) is positive and x-\left(5-\sqrt{33}\right) is negative.
x\in \emptyset
This is false for any x.
x-\left(5-\sqrt{33}\right)>0 x-\left(\sqrt{33}+5\right)<0
Consider the case when x-\left(5-\sqrt{33}\right) is positive and x-\left(\sqrt{33}+5\right) is negative.
x\in \left(5-\sqrt{33},\sqrt{33}+5\right)
The solution satisfying both inequalities is x\in \left(5-\sqrt{33},\sqrt{33}+5\right).
x\in \left(5-\sqrt{33},\sqrt{33}+5\right)
The final solution is the union of the obtained solutions.