Solve t^2+8t+16=45

$$t ^ { 2 } + 8 t + 16 = 45$$

Solve for t

$$t=3\sqrt{5}-4\approx 2.708203932$$

$$t=-3\sqrt{5}-4\approx -10.708203932$$

Steps Using the Quadratic Formula

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.

$$t^{2}+8t+16=45$$

Subtract 45 from both sides of the equation.

$$t^{2}+8t+16-45=45-45$$

Subtracting 45 from itself leaves 0.

$$t^{2}+8t+16-45=0$$

Subtract 45 from 16.

$$t^{2}+8t-29=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 8 for b, and -29 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-8±\sqrt{8^{2}-4\left(-29\right)}}{2}$$

Square 8.

$$t=\frac{-8±\sqrt{64-4\left(-29\right)}}{2}$$

Multiply -4 times -29.

$$t=\frac{-8±\sqrt{64+116}}{2}$$

Add 64 to 116.

$$t=\frac{-8±\sqrt{180}}{2}$$

Take the square root of 180.

$$t=\frac{-8±6\sqrt{5}}{2}$$

Now solve the equation t=\frac{-8±6\sqrt{5}}{2} when ± is plus. Add -8 to 6\sqrt{5}.

$$t=\frac{6\sqrt{5}-8}{2}$$

Divide -8+6\sqrt{5} by 2.

$$t=3\sqrt{5}-4$$

Now solve the equation t=\frac{-8±6\sqrt{5}}{2} when ± is minus. Subtract 6\sqrt{5} from -8.

$$t=\frac{-6\sqrt{5}-8}{2}$$

Divide -8-6\sqrt{5} by 2.

$$t=-3\sqrt{5}-4$$

The equation is now solved.

$$t=3\sqrt{5}-4 t=-3\sqrt{5}-4$$

Steps for Completing the Square

Factor t^{2}+8t+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t+4\right)^{2}=45$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t+4\right)^{2}}=\sqrt{45}$$

Simplify.

$$t+4=3\sqrt{5} t+4=-3\sqrt{5}$$

Subtract 4 from both sides of the equation.

$$t=3\sqrt{5}-4 t=-3\sqrt{5}-4$$