Solve t^2:2^4-6t-2^8=0

$$t ^ { 2 } : 2 ^ { 4 } - 6 t - 2 ^ { 8 } = 0$$

Solve for t

$$t=-32$$

$$t=128$$

Steps Using Factoring

Calculate 2 to the power of 4 and get 16.

$$\frac{t^{2}}{16}-6t-2^{8}=0$$

Calculate 2 to the power of 8 and get 256.

$$\frac{t^{2}}{16}-6t-256=0$$

Multiply both sides of the equation by 16.

$$t^{2}-96t-4096=0$$

To solve the equation, factor t^{2}-96t-4096 using formula t^{2}+\left(a+b\right)t+ab=\left(t+a\right)\left(t+b\right). To find a and b, set up a system to be solved.

$$a+b=-96 ab=-4096$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -4096.

$$1,-4096 2,-2048 4,-1024 8,-512 16,-256 32,-128 64,-64$$

Calculate the sum for each pair.

$$1-4096=-4095 2-2048=-2046 4-1024=-1020 8-512=-504 16-256=-240 32-128=-96 64-64=0$$

The solution is the pair that gives sum -96.

$$a=-128 b=32$$

Rewrite factored expression \left(t+a\right)\left(t+b\right) using the obtained values.

$$\left(t-128\right)\left(t+32\right)$$

To find equation solutions, solve t-128=0 and t+32=0.

$$t=128 t=-32$$

Steps Using Factoring By Grouping

Calculate 2 to the power of 4 and get 16.

$$\frac{t^{2}}{16}-6t-2^{8}=0$$

Calculate 2 to the power of 8 and get 256.

$$\frac{t^{2}}{16}-6t-256=0$$

Multiply both sides of the equation by 16.

$$t^{2}-96t-4096=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as t^{2}+at+bt-4096. To find a and b, set up a system to be solved.

$$a+b=-96 ab=1\left(-4096\right)=-4096$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -4096.

$$1,-4096 2,-2048 4,-1024 8,-512 16,-256 32,-128 64,-64$$

Calculate the sum for each pair.

$$1-4096=-4095 2-2048=-2046 4-1024=-1020 8-512=-504 16-256=-240 32-128=-96 64-64=0$$

The solution is the pair that gives sum -96.

$$a=-128 b=32$$

Rewrite t^{2}-96t-4096 as \left(t^{2}-128t\right)+\left(32t-4096\right).

$$\left(t^{2}-128t\right)+\left(32t-4096\right)$$

Factor out t in the first and 32 in the second group.

$$t\left(t-128\right)+32\left(t-128\right)$$

Factor out common term t-128 by using distributive property.

$$\left(t-128\right)\left(t+32\right)$$

To find equation solutions, solve t-128=0 and t+32=0.

$$t=128 t=-32$$

Steps Using the Quadratic Formula

Calculate 2 to the power of 4 and get 16.

$$\frac{t^{2}}{16}-6t-2^{8}=0$$

Calculate 2 to the power of 8 and get 256.

$$\frac{t^{2}}{16}-6t-256=0$$

Multiply both sides of the equation by 16.

$$t^{2}-96t-4096=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -96 for b, and -4096 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-96\right)±\sqrt{\left(-96\right)^{2}-4\left(-4096\right)}}{2}$$

Square -96.

$$t=\frac{-\left(-96\right)±\sqrt{9216-4\left(-4096\right)}}{2}$$

Multiply -4 times -4096.

$$t=\frac{-\left(-96\right)±\sqrt{9216+16384}}{2}$$

Add 9216 to 16384.

$$t=\frac{-\left(-96\right)±\sqrt{25600}}{2}$$

Take the square root of 25600.

$$t=\frac{-\left(-96\right)±160}{2}$$

The opposite of -96 is 96.

$$t=\frac{96±160}{2}$$

Now solve the equation t=\frac{96±160}{2} when ± is plus. Add 96 to 160.

$$t=\frac{256}{2}$$

Divide 256 by 2.

$$t=128$$

Now solve the equation t=\frac{96±160}{2} when ± is minus. Subtract 160 from 96.

$$t=-\frac{64}{2}$$

Divide -64 by 2.

$$t=-32$$

The equation is now solved.

$$t=128 t=-32$$

Steps for Completing the Square

Calculate 2 to the power of 4 and get 16.

$$\frac{t^{2}}{16}-6t-2^{8}=0$$

Calculate 2 to the power of 8 and get 256.

$$\frac{t^{2}}{16}-6t-256=0$$

Add 256 to both sides. Anything plus zero gives itself.

$$\frac{t^{2}}{16}-6t=256$$

Multiply both sides of the equation by 16.

$$t^{2}-96t=4096$$

Divide -96, the coefficient of the x term, by 2 to get -48. Then add the square of -48 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-96t+\left(-48\right)^{2}=4096+\left(-48\right)^{2}$$

Square -48.

$$t^{2}-96t+2304=4096+2304$$

Add 4096 to 2304.

$$t^{2}-96t+2304=6400$$

Factor t^{2}-96t+2304. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-48\right)^{2}=6400$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-48\right)^{2}}=\sqrt{6400}$$

Simplify.

$$t-48=80 t-48=-80$$

Add 48 to both sides of the equation.

$$t=128 t=-32$$