Solve 4n^2-n-812

$$4 n ^ { 2 } - n - 812$$

Factor

$$4\left(n-\frac{1-\sqrt{12993}}{8}\right)\left(n-\frac{\sqrt{12993}+1}{8}\right)$$

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.

$$4n^{2}-n-812=0$$

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.

$$n=\frac{-\left(-1\right)±\sqrt{1-4\times 4\left(-812\right)}}{2\times 4}$$

Multiply -4 times 4.

$$n=\frac{-\left(-1\right)±\sqrt{1-16\left(-812\right)}}{2\times 4}$$

Multiply -16 times -812.

$$n=\frac{-\left(-1\right)±\sqrt{1+12992}}{2\times 4}$$

Add 1 to 12992.

$$n=\frac{-\left(-1\right)±\sqrt{12993}}{2\times 4}$$

The opposite of -1 is 1.

$$n=\frac{1±\sqrt{12993}}{2\times 4}$$

Multiply 2 times 4.

$$n=\frac{1±\sqrt{12993}}{8}$$

Now solve the equation n=\frac{1±\sqrt{12993}}{8} when ± is plus. Add 1 to \sqrt{12993}.

$$n=\frac{\sqrt{12993}+1}{8}$$

Now solve the equation n=\frac{1±\sqrt{12993}}{8} when ± is minus. Subtract \sqrt{12993} from 1.

$$n=\frac{1-\sqrt{12993}}{8}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1+\sqrt{12993}}{8} for x_{1} and \frac{1-\sqrt{12993}}{8} for x_{2}.

$$4n^{2}-n-812=4\left(n-\frac{\sqrt{12993}+1}{8}\right)\left(n-\frac{1-\sqrt{12993}}{8}\right)$$

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.This is achieved by dividing both sides of the equation by 4

$$x ^ 2 -\frac{1}{4}x -203 = 0$$

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

$$r + s = \frac{1}{4} rs = -203$$

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

$$r = \frac{1}{8} - u s = \frac{1}{8} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -203

$$(\frac{1}{8} - u) (\frac{1}{8} + u) = -203$$

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

$$\frac{1}{64} - u^2 = -203$$

Simplify the expression by subtracting \frac{1}{64} on both sides

$$-u^2 = -203-\frac{1}{64} = -\frac{12993}{64}$$

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

$$u^2 = \frac{12993}{64} u = \pm\sqrt{\frac{12993}{64}} = \pm \frac{\sqrt{12993}}{8}$$

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

$$r =\frac{1}{8} - \frac{\sqrt{12993}}{8} = -14.123 s = \frac{1}{8} + \frac{\sqrt{12993}}{8} = 14.373$$

Evaluate

$$4n^{2}-n-812$$