Solve b^2+2b-47>0

$$b ^ { 2 } + 2 b - 47 > 0$$

Solve for b

$$b\in \left(-\infty,-4\sqrt{3}-1\right)\cup \left(4\sqrt{3}-1,\infty\right)$$

Steps Using the Quadratic Formula

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.

$$b^{2}+2b-47=0$$

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, 2 for b, and -47 for c in the quadratic formula.

$$b=\frac{-2±\sqrt{2^{2}-4\times 1\left(-47\right)}}{2}$$

Do the calculations.

$$b=\frac{-2±8\sqrt{3}}{2}$$

Solve the equation b=\frac{-2±8\sqrt{3}}{2} when ± is plus and when ± is minus.

$$b=4\sqrt{3}-1 b=-4\sqrt{3}-1$$

Rewrite the inequality by using the obtained solutions.

$$\left(b-\left(4\sqrt{3}-1\right)\right)\left(b-\left(-4\sqrt{3}-1\right)\right)>0$$

For the product to be positive, b-\left(4\sqrt{3}-1\right) and b-\left(-4\sqrt{3}-1\right) have to be both negative or both positive. Consider the case when b-\left(4\sqrt{3}-1\right) and b-\left(-4\sqrt{3}-1\right) are both negative.

$$b-\left(4\sqrt{3}-1\right)<0 b-\left(-4\sqrt{3}-1\right)<0$$

The solution satisfying both inequalities is b<-4\sqrt{3}-1.

$$b<-4\sqrt{3}-1$$

Consider the case when b-\left(4\sqrt{3}-1\right) and b-\left(-4\sqrt{3}-1\right) are both positive.

$$b-\left(-4\sqrt{3}-1\right)>0 b-\left(4\sqrt{3}-1\right)>0$$

The solution satisfying both inequalities is b>4\sqrt{3}-1.

$$b>4\sqrt{3}-1$$

The final solution is the union of the obtained solutions.

$$b<-4\sqrt{3}-1\text{; }b>4\sqrt{3}-1$$