Solve b^2+70>16b

$$b ^ { 2 } + 70 > 16 b$$

Solve for b

$$b\in \mathrm{R}$$

Steps Using the Quadratic Formula

Subtract 16b from both sides.

$$b^{2}+70-16b>0$$

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.

$$b^{2}+70-16b=0$$

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, -16 for b, and 70 for c in the quadratic formula.

$$b=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 1\times 70}}{2}$$

Do the calculations.

$$b=\frac{16±\sqrt{-24}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions. Expression b^{2}+70-16b has the same sign for any b. To determine the sign, calculate the value of the expression for b=0.

$$0^{2}+70-16\times 0=70$$

The value of the expression b^{2}+70-16b is always positive. Inequality holds for b\in \mathrm{R}.

$$b\in \mathrm{R}$$