Solve 16t+t^2+14

$$16 t + t ^ { 2 } + 14$$

Factor

$$\left(t-\left(-5\sqrt{2}-8\right)\right)\left(t-\left(5\sqrt{2}-8\right)\right)$$

Steps Using the Quadratic Formula

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.

$$t^{2}+16t+14=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$t=\frac{-16±\sqrt{16^{2}-4\times 14}}{2}$$

Square 16.

$$t=\frac{-16±\sqrt{256-4\times 14}}{2}$$

Multiply -4 times 14.

$$t=\frac{-16±\sqrt{256-56}}{2}$$

Add 256 to -56.

$$t=\frac{-16±\sqrt{200}}{2}$$

Take the square root of 200.

$$t=\frac{-16±10\sqrt{2}}{2}$$

Now solve the equation t=\frac{-16±10\sqrt{2}}{2} when ± is plus. Add -16 to 10\sqrt{2}.

$$t=\frac{10\sqrt{2}-16}{2}$$

Divide -16+10\sqrt{2} by 2.

$$t=5\sqrt{2}-8$$

Now solve the equation t=\frac{-16±10\sqrt{2}}{2} when ± is minus. Subtract 10\sqrt{2} from -16.

$$t=\frac{-10\sqrt{2}-16}{2}$$

Divide -16-10\sqrt{2} by 2.

$$t=-5\sqrt{2}-8$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -8+5\sqrt{2} for x_{1} and -8-5\sqrt{2} for x_{2}.

$$t^{2}+16t+14=\left(t-\left(5\sqrt{2}-8\right)\right)\left(t-\left(-5\sqrt{2}-8\right)\right)$$

Evaluate

$$t^{2}+16t+14$$