Solve -16t^2+88t+4

$$- 16 t ^ { 2 } + 88 t + 4$$

Factor

$$-16\left(t-\frac{11-5\sqrt{5}}{4}\right)\left(t-\frac{5\sqrt{5}+11}{4}\right)$$

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.

$$-16t^{2}+88t+4=0$$

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.

$$t=\frac{-88±\sqrt{88^{2}-4\left(-16\right)\times 4}}{2\left(-16\right)}$$

Square 88.

$$t=\frac{-88±\sqrt{7744-4\left(-16\right)\times 4}}{2\left(-16\right)}$$

Multiply -4 times -16.

$$t=\frac{-88±\sqrt{7744+64\times 4}}{2\left(-16\right)}$$

Multiply 64 times 4.

$$t=\frac{-88±\sqrt{7744+256}}{2\left(-16\right)}$$

Add 7744 to 256.

$$t=\frac{-88±\sqrt{8000}}{2\left(-16\right)}$$

Take the square root of 8000.

$$t=\frac{-88±40\sqrt{5}}{2\left(-16\right)}$$

Multiply 2 times -16.

$$t=\frac{-88±40\sqrt{5}}{-32}$$

Now solve the equation t=\frac{-88±40\sqrt{5}}{-32} when ± is plus. Add -88 to 40\sqrt{5}.

$$t=\frac{40\sqrt{5}-88}{-32}$$

Divide -88+40\sqrt{5} by -32.

$$t=\frac{11-5\sqrt{5}}{4}$$

Now solve the equation t=\frac{-88±40\sqrt{5}}{-32} when ± is minus. Subtract 40\sqrt{5} from -88.

$$t=\frac{-40\sqrt{5}-88}{-32}$$

Divide -88-40\sqrt{5} by -32.

$$t=\frac{5\sqrt{5}+11}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{11-5\sqrt{5}}{4} for x_{1} and \frac{11+5\sqrt{5}}{4} for x_{2}.

$$-16t^{2}+88t+4=-16\left(t-\frac{11-5\sqrt{5}}{4}\right)\left(t-\frac{5\sqrt{5}+11}{4}\right)$$

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.

$$x ^ 2 -\frac{11}{2}x -\frac{1}{4} = 0$$

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

$$r + s = \frac{11}{2} rs = -\frac{1}{4}$$

Two numbers r and s sum up to \frac{11}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{11}{2} = \frac{11}{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>

$$r = \frac{11}{4} - u s = \frac{11}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{4}

$$(\frac{11}{4} - u) (\frac{11}{4} + u) = -\frac{1}{4}$$

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

$$\frac{121}{16} - u^2 = -\frac{1}{4}$$

Simplify the expression by subtracting \frac{121}{16} on both sides

$$-u^2 = -\frac{1}{4}-\frac{121}{16} = -\frac{125}{16}$$

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.

$$r =\frac{11}{4} - \frac{\sqrt{125}}{4} = -0.045 s = \frac{11}{4} + \frac{\sqrt{125}}{4} = 5.545$$

Evaluate

$$4+88t-16t^{2}$$