Solve h(t)=-36t^2+96t+112

$$h ( t ) = - 36 t ^ { 2 } + 96 t + 112$$

Factor

$$-36\left(t-\frac{4-2\sqrt{11}}{3}\right)\left(t-\frac{2\sqrt{11}+4}{3}\right)$$

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.

$$-36t^{2}+96t+112=0$$

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.

$$t=\frac{-96±\sqrt{96^{2}-4\left(-36\right)\times 112}}{2\left(-36\right)}$$

Square 96.

$$t=\frac{-96±\sqrt{9216-4\left(-36\right)\times 112}}{2\left(-36\right)}$$

Multiply -4 times -36.

$$t=\frac{-96±\sqrt{9216+144\times 112}}{2\left(-36\right)}$$

Multiply 144 times 112.

$$t=\frac{-96±\sqrt{9216+16128}}{2\left(-36\right)}$$

Add 9216 to 16128.

$$t=\frac{-96±\sqrt{25344}}{2\left(-36\right)}$$

Take the square root of 25344.

$$t=\frac{-96±48\sqrt{11}}{2\left(-36\right)}$$

Multiply 2 times -36.

$$t=\frac{-96±48\sqrt{11}}{-72}$$

Now solve the equation t=\frac{-96±48\sqrt{11}}{-72} when ± is plus. Add -96 to 48\sqrt{11}.

$$t=\frac{48\sqrt{11}-96}{-72}$$

Divide -96+48\sqrt{11} by -72.

$$t=\frac{4-2\sqrt{11}}{3}$$

Now solve the equation t=\frac{-96±48\sqrt{11}}{-72} when ± is minus. Subtract 48\sqrt{11} from -96.

$$t=\frac{-48\sqrt{11}-96}{-72}$$

Divide -96-48\sqrt{11} by -72.

$$t=\frac{2\sqrt{11}+4}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{4-2\sqrt{11}}{3} for x_{1} and \frac{4+2\sqrt{11}}{3} for x_{2}.

$$-36t^{2}+96t+112=-36\left(t-\frac{4-2\sqrt{11}}{3}\right)\left(t-\frac{2\sqrt{11}+4}{3}\right)$$

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.

$$x ^ 2 -\frac{8}{3}x -\frac{28}{9} = 0$$

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

$$r + s = \frac{8}{3} rs = -\frac{28}{9}$$

Two numbers r and s sum up to \frac{8}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{8}{3} = \frac{4}{3}. 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>

$$r = \frac{4}{3} - u s = \frac{4}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{28}{9}

$$(\frac{4}{3} - u) (\frac{4}{3} + u) = -\frac{28}{9}$$

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

$$\frac{16}{9} - u^2 = -\frac{28}{9}$$

Simplify the expression by subtracting \frac{16}{9} on both sides

$$-u^2 = -\frac{28}{9}-\frac{16}{9} = -\frac{44}{9}$$

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

$$u^2 = \frac{44}{9} u = \pm\sqrt{\frac{44}{9}} = \pm \frac{\sqrt{44}}{3}$$

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

$$r =\frac{4}{3} - \frac{\sqrt{44}}{3} = -0.878 s = \frac{4}{3} + \frac{\sqrt{44}}{3} = 3.544$$

Evaluate

$$112+96t-36t^{2}$$