Solve -16t^2+64t+80=128

$$- 16 t ^ { 2 } + 64 t + 80 = 128$$

Solve for t

$$t=1$$

$$t=3$$

Steps Using Factoring By Grouping

Subtract 128 from both sides.

$$-16t^{2}+64t+80-128=0$$

Subtract 128 from 80 to get -48.

$$-16t^{2}+64t-48=0$$

Divide both sides by 16.

$$-t^{2}+4t-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -t^{2}+at+bt-3. To find a and b, set up a system to be solved.

$$a+b=4 ab=-\left(-3\right)=3$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=3 b=1$$

Rewrite -t^{2}+4t-3 as \left(-t^{2}+3t\right)+\left(t-3\right).

$$\left(-t^{2}+3t\right)+\left(t-3\right)$$

Factor out -t in -t^{2}+3t.

$$-t\left(t-3\right)+t-3$$

Factor out common term t-3 by using distributive property.

$$\left(t-3\right)\left(-t+1\right)$$

To find equation solutions, solve t-3=0 and -t+1=0.

$$t=3 t=1$$

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.

$$-16t^{2}+64t+80=128$$

Subtract 128 from both sides of the equation.

$$-16t^{2}+64t+80-128=128-128$$

Subtracting 128 from itself leaves 0.

$$-16t^{2}+64t+80-128=0$$

Subtract 128 from 80.

$$-16t^{2}+64t-48=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -16 for a, 64 for b, and -48 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-64±\sqrt{64^{2}-4\left(-16\right)\left(-48\right)}}{2\left(-16\right)}$$

Square 64.

$$t=\frac{-64±\sqrt{4096-4\left(-16\right)\left(-48\right)}}{2\left(-16\right)}$$

Multiply -4 times -16.

$$t=\frac{-64±\sqrt{4096+64\left(-48\right)}}{2\left(-16\right)}$$

Multiply 64 times -48.

$$t=\frac{-64±\sqrt{4096-3072}}{2\left(-16\right)}$$

Add 4096 to -3072.

$$t=\frac{-64±\sqrt{1024}}{2\left(-16\right)}$$

Take the square root of 1024.

$$t=\frac{-64±32}{2\left(-16\right)}$$

Multiply 2 times -16.

$$t=\frac{-64±32}{-32}$$

Now solve the equation t=\frac{-64±32}{-32} when ± is plus. Add -64 to 32.

$$t=-\frac{32}{-32}$$

Divide -32 by -32.

$$t=1$$

Now solve the equation t=\frac{-64±32}{-32} when ± is minus. Subtract 32 from -64.

$$t=-\frac{96}{-32}$$

Divide -96 by -32.

$$t=3$$

The equation is now solved.

$$t=1 t=3$$

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-16t^{2}+64t+80=128$$

Subtract 80 from both sides of the equation.

$$-16t^{2}+64t+80-80=128-80$$

Subtracting 80 from itself leaves 0.

$$-16t^{2}+64t=128-80$$

Subtract 80 from 128.

$$-16t^{2}+64t=48$$

Divide both sides by -16.

$$\frac{-16t^{2}+64t}{-16}=\frac{48}{-16}$$

Dividing by -16 undoes the multiplication by -16.

$$t^{2}+\frac{64}{-16}t=\frac{48}{-16}$$

Divide 64 by -16.

$$t^{2}-4t=\frac{48}{-16}$$

Divide 48 by -16.

$$t^{2}-4t=-3$$

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-4t+\left(-2\right)^{2}=-3+\left(-2\right)^{2}$$

Square -2.

$$t^{2}-4t+4=-3+4$$

Add -3 to 4.

$$t^{2}-4t+4=1$$

Factor t^{2}-4t+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-2\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-2\right)^{2}}=\sqrt{1}$$

Simplify.

$$t-2=1 t-2=-1$$

Add 2 to both sides of the equation.

$$t=3 t=1$$