Solve h(t)=-16t^2+40t+6

$$h ( t ) = - 16 t ^ { 2 } + 40 t + 6$$

Factor

$$-16\left(t-\frac{5-\sqrt{31}}{4}\right)\left(t-\frac{\sqrt{31}+5}{4}\right)$$

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.

$$-16t^{2}+40t+6=0$$

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.

$$t=\frac{-40±\sqrt{40^{2}-4\left(-16\right)\times 6}}{2\left(-16\right)}$$

Square 40.

$$t=\frac{-40±\sqrt{1600-4\left(-16\right)\times 6}}{2\left(-16\right)}$$

Multiply -4 times -16.

$$t=\frac{-40±\sqrt{1600+64\times 6}}{2\left(-16\right)}$$

Multiply 64 times 6.

$$t=\frac{-40±\sqrt{1600+384}}{2\left(-16\right)}$$

Add 1600 to 384.

$$t=\frac{-40±\sqrt{1984}}{2\left(-16\right)}$$

Take the square root of 1984.

$$t=\frac{-40±8\sqrt{31}}{2\left(-16\right)}$$

Multiply 2 times -16.

$$t=\frac{-40±8\sqrt{31}}{-32}$$

Now solve the equation t=\frac{-40±8\sqrt{31}}{-32} when ± is plus. Add -40 to 8\sqrt{31}.

$$t=\frac{8\sqrt{31}-40}{-32}$$

Divide -40+8\sqrt{31} by -32.

$$t=\frac{5-\sqrt{31}}{4}$$

Now solve the equation t=\frac{-40±8\sqrt{31}}{-32} when ± is minus. Subtract 8\sqrt{31} from -40.

$$t=\frac{-8\sqrt{31}-40}{-32}$$

Divide -40-8\sqrt{31} by -32.

$$t=\frac{\sqrt{31}+5}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5-\sqrt{31}}{4} for x_{1} and \frac{5+\sqrt{31}}{4} for x_{2}.

$$-16t^{2}+40t+6=-16\left(t-\frac{5-\sqrt{31}}{4}\right)\left(t-\frac{\sqrt{31}+5}{4}\right)$$

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.

$$x ^ 2 -\frac{5}{2}x -\frac{3}{8} = 0$$

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

$$r + s = \frac{5}{2} rs = -\frac{3}{8}$$

Two numbers r and s sum up to \frac{5}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{5}{2} = \frac{5}{4}. 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>

$$r = \frac{5}{4} - u s = \frac{5}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{8}

$$(\frac{5}{4} - u) (\frac{5}{4} + u) = -\frac{3}{8}$$

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

$$\frac{25}{16} - u^2 = -\frac{3}{8}$$

Simplify the expression by subtracting \frac{25}{16} on both sides

$$-u^2 = -\frac{3}{8}-\frac{25}{16} = -\frac{31}{16}$$

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

$$u^2 = \frac{31}{16} u = \pm\sqrt{\frac{31}{16}} = \pm \frac{\sqrt{31}}{4}$$

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

$$r =\frac{5}{4} - \frac{\sqrt{31}}{4} = -0.142 s = \frac{5}{4} + \frac{\sqrt{31}}{4} = 2.642$$

Evaluate

$$6+40t-16t^{2}$$