Solve s(t)=-16t^2+64t+80

$$s ( t ) = - 16 t ^ { 2 } + 64 t + 80$$

Factor

$$-16\left(t-5\right)\left(t+1\right)$$

Solution Steps

Factor out 16.

$$16\left(-t^{2}+4t+5\right)$$

Consider -t^{2}+4t+5. Factor the expression by grouping. First, the expression needs to be rewritten as -t^{2}+at+bt+5. To find a and b, set up a system to be solved.

$$a+b=4 ab=-5=-5$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=5 b=-1$$

Rewrite -t^{2}+4t+5 as \left(-t^{2}+5t\right)+\left(-t+5\right).

$$\left(-t^{2}+5t\right)+\left(-t+5\right)$$

Factor out -t in the first and -1 in the second group.

$$-t\left(t-5\right)-\left(t-5\right)$$

Factor out common term t-5 by using distributive property.

$$\left(t-5\right)\left(-t-1\right)$$

Rewrite the complete factored expression.

$$16\left(t-5\right)\left(-t-1\right)$$

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.

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

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.

$$t=\frac{-64±\sqrt{64^{2}-4\left(-16\right)\times 80}}{2\left(-16\right)}$$

Square 64.

$$t=\frac{-64±\sqrt{4096-4\left(-16\right)\times 80}}{2\left(-16\right)}$$

Multiply -4 times -16.

$$t=\frac{-64±\sqrt{4096+64\times 80}}{2\left(-16\right)}$$

Multiply 64 times 80.

$$t=\frac{-64±\sqrt{4096+5120}}{2\left(-16\right)}$$

Add 4096 to 5120.

$$t=\frac{-64±\sqrt{9216}}{2\left(-16\right)}$$

Take the square root of 9216.

$$t=\frac{-64±96}{2\left(-16\right)}$$

Multiply 2 times -16.

$$t=\frac{-64±96}{-32}$$

Now solve the equation t=\frac{-64±96}{-32} when ± is plus. Add -64 to 96.

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

Divide 32 by -32.

$$t=-1$$

Now solve the equation t=\frac{-64±96}{-32} when ± is minus. Subtract 96 from -64.

$$t=-\frac{160}{-32}$$

Divide -160 by -32.

$$t=5$$

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

$$-16t^{2}+64t+80=-16\left(t-\left(-1\right)\right)\left(t-5\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$-16t^{2}+64t+80=-16\left(t+1\right)\left(t-5\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.

$$x ^ 2 -4x -5 = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = 4 rs = -5$$

Two numbers r and s sum up to 4 exactly when the average of the two numbers is \frac{1}{2}*4 = 2. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = 2 - u s = 2 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -5

$$(2 - u) (2 + u) = -5$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$4 - u^2 = -5$$

Simplify the expression by subtracting 4 on both sides

$$-u^2 = -5-4 = -9$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = 9 u = \pm\sqrt{9} = \pm 3$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =2 - 3 = -1 s = 2 + 3 = 5$$

Evaluate

$$-16\left(t-5\right)\left(t+1\right)$$