Solve h(t)=-5t^2+5t+5

$$h ( t ) = - 5 t ^ { 2 } + 5 t + 5$$

Factor

$$-5\left(t-\frac{1-\sqrt{5}}{2}\right)\left(t-\frac{\sqrt{5}+1}{2}\right)$$

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.

$$-5t^{2}+5t+5=0$$

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.

$$t=\frac{-5±\sqrt{5^{2}-4\left(-5\right)\times 5}}{2\left(-5\right)}$$

Square 5.

$$t=\frac{-5±\sqrt{25-4\left(-5\right)\times 5}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$t=\frac{-5±\sqrt{25+20\times 5}}{2\left(-5\right)}$$

Multiply 20 times 5.

$$t=\frac{-5±\sqrt{25+100}}{2\left(-5\right)}$$

Add 25 to 100.

$$t=\frac{-5±\sqrt{125}}{2\left(-5\right)}$$

Take the square root of 125.

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

Multiply 2 times -5.

$$t=\frac{-5±5\sqrt{5}}{-10}$$

Now solve the equation t=\frac{-5±5\sqrt{5}}{-10} when ± is plus. Add -5 to 5\sqrt{5}.

$$t=\frac{5\sqrt{5}-5}{-10}$$

Divide -5+5\sqrt{5} by -10.

$$t=\frac{1-\sqrt{5}}{2}$$

Now solve the equation t=\frac{-5±5\sqrt{5}}{-10} when ± is minus. Subtract 5\sqrt{5} from -5.

$$t=\frac{-5\sqrt{5}-5}{-10}$$

Divide -5-5\sqrt{5} by -10.

$$t=\frac{\sqrt{5}+1}{2}$$

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

$$-5t^{2}+5t+5=-5\left(t-\frac{1-\sqrt{5}}{2}\right)\left(t-\frac{\sqrt{5}+1}{2}\right)$$

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.

$$x ^ 2 -1x -1 = 0$$

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

$$r + s = 1 rs = -1$$

Two numbers r and s sum up to 1 exactly when the average of the two numbers is \frac{1}{2}*1 = \frac{1}{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>

$$r = \frac{1}{2} - u s = \frac{1}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -1

$$(\frac{1}{2} - u) (\frac{1}{2} + u) = -1$$

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

$$\frac{1}{4} - u^2 = -1$$

Simplify the expression by subtracting \frac{1}{4} on both sides

$$-u^2 = -1-\frac{1}{4} = -\frac{5}{4}$$

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

$$u^2 = \frac{5}{4} u = \pm\sqrt{\frac{5}{4}} = \pm \frac{\sqrt{5}}{2}$$

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

$$r =\frac{1}{2} - \frac{\sqrt{5}}{2} = -0.618 s = \frac{1}{2} + \frac{\sqrt{5}}{2} = 1.618$$

Evaluate

$$5\left(1+t-t^{2}\right)$$