Solve h(t)=-16t^2+80t+25

$$h ( t ) = - 16 t ^ { 2 } + 80 t + 25$$

Factor

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

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.

$$-16t^{2}+80t+25=0$$

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.

$$t=\frac{-80±\sqrt{80^{2}-4\left(-16\right)\times 25}}{2\left(-16\right)}$$

Square 80.

$$t=\frac{-80±\sqrt{6400-4\left(-16\right)\times 25}}{2\left(-16\right)}$$

Multiply -4 times -16.

$$t=\frac{-80±\sqrt{6400+64\times 25}}{2\left(-16\right)}$$

Multiply 64 times 25.

$$t=\frac{-80±\sqrt{6400+1600}}{2\left(-16\right)}$$

Add 6400 to 1600.

$$t=\frac{-80±\sqrt{8000}}{2\left(-16\right)}$$

Take the square root of 8000.

$$t=\frac{-80±40\sqrt{5}}{2\left(-16\right)}$$

Multiply 2 times -16.

$$t=\frac{-80±40\sqrt{5}}{-32}$$

Now solve the equation t=\frac{-80±40\sqrt{5}}{-32} when ± is plus. Add -80 to 40\sqrt{5}.

$$t=\frac{40\sqrt{5}-80}{-32}$$

Divide -80+40\sqrt{5} by -32.

$$t=-\frac{5\sqrt{5}}{4}+\frac{5}{2}$$

Now solve the equation t=\frac{-80±40\sqrt{5}}{-32} when ± is minus. Subtract 40\sqrt{5} from -80.

$$t=\frac{-40\sqrt{5}-80}{-32}$$

Divide -80-40\sqrt{5} by -32.

$$t=\frac{5\sqrt{5}}{4}+\frac{5}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{2}-\frac{5\sqrt{5}}{4} for x_{1} and \frac{5}{2}+\frac{5\sqrt{5}}{4} for x_{2}.

$$-16t^{2}+80t+25=-16\left(t-\left(-\frac{5\sqrt{5}}{4}+\frac{5}{2}\right)\right)\left(t-\left(\frac{5\sqrt{5}}{4}+\frac{5}{2}\right)\right)$$

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.

$$x ^ 2 -5x -\frac{25}{16} = 0$$

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

$$r + s = 5 rs = -\frac{25}{16}$$

Two numbers r and s sum up to 5 exactly when the average of the two numbers is \frac{1}{2}*5 = \frac{5}{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-gzdabgg4ehffg0hf.b01.azurefd.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{5}{2} - u s = \frac{5}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{25}{16}

$$(\frac{5}{2} - u) (\frac{5}{2} + u) = -\frac{25}{16}$$

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

$$\frac{25}{4} - u^2 = -\frac{25}{16}$$

Simplify the expression by subtracting \frac{25}{4} on both sides

$$-u^2 = -\frac{25}{16}-\frac{25}{4} = -\frac{125}{16}$$

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

$$u^2 = \frac{125}{16} u = \pm\sqrt{\frac{125}{16}} = \pm \frac{\sqrt{125}}{4}$$

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

$$r =\frac{5}{2} - \frac{\sqrt{125}}{4} = -0.295 s = \frac{5}{2} + \frac{\sqrt{125}}{4} = 5.295$$

Evaluate

$$25+80t-16t^{2}$$