Solve 2n^2+12n+13

$$2 n ^ { 2 } + 12 n + 13$$

Factor

$$2\left(n-\left(-\frac{\sqrt{10}}{2}-3\right)\right)\left(n-\left(\frac{\sqrt{10}}{2}-3\right)\right)$$

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.

$$2n^{2}+12n+13=0$$

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.

$$n=\frac{-12±\sqrt{12^{2}-4\times 2\times 13}}{2\times 2}$$

Square 12.

$$n=\frac{-12±\sqrt{144-4\times 2\times 13}}{2\times 2}$$

Multiply -4 times 2.

$$n=\frac{-12±\sqrt{144-8\times 13}}{2\times 2}$$

Multiply -8 times 13.

$$n=\frac{-12±\sqrt{144-104}}{2\times 2}$$

Add 144 to -104.

$$n=\frac{-12±\sqrt{40}}{2\times 2}$$

Take the square root of 40.

$$n=\frac{-12±2\sqrt{10}}{2\times 2}$$

Multiply 2 times 2.

$$n=\frac{-12±2\sqrt{10}}{4}$$

Now solve the equation n=\frac{-12±2\sqrt{10}}{4} when ± is plus. Add -12 to 2\sqrt{10}.

$$n=\frac{2\sqrt{10}-12}{4}$$

Divide -12+2\sqrt{10} by 4.

$$n=\frac{\sqrt{10}}{2}-3$$

Now solve the equation n=\frac{-12±2\sqrt{10}}{4} when ± is minus. Subtract 2\sqrt{10} from -12.

$$n=\frac{-2\sqrt{10}-12}{4}$$

Divide -12-2\sqrt{10} by 4.

$$n=-\frac{\sqrt{10}}{2}-3$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -3+\frac{\sqrt{10}}{2} for x_{1} and -3-\frac{\sqrt{10}}{2} for x_{2}.

$$2n^{2}+12n+13=2\left(n-\left(\frac{\sqrt{10}}{2}-3\right)\right)\left(n-\left(-\frac{\sqrt{10}}{2}-3\right)\right)$$

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 2

$$x ^ 2 +6x +\frac{13}{2} = 0$$

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

$$r + s = -6 rs = \frac{13}{2}$$

Two numbers r and s sum up to -6 exactly when the average of the two numbers is \frac{1}{2}*-6 = -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>

$$r = -3 - u s = -3 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{13}{2}

$$(-3 - u) (-3 + u) = \frac{13}{2}$$

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

$$9 - u^2 = \frac{13}{2}$$

Simplify the expression by subtracting 9 on both sides

$$-u^2 = \frac{13}{2}-9 = -\frac{5}{2}$$

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

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

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

$$r =-3 - \frac{\sqrt{5}}{\sqrt{2}} = -4.581 s = -3 + \frac{\sqrt{5}}{\sqrt{2}} = -1.419$$

Evaluate

$$2n^{2}+12n+13$$