Solve {l}{2x^2+20x}{+6}

$$\left. \begin{array} { l } { 2 x ^ { 2 } + 20 x } \\ { + 6 } \end{array} \right.$$

Factor

$$2\left(x-\left(-\sqrt{22}-5\right)\right)\left(x-\left(\sqrt{22}-5\right)\right)$$

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.

$$2x^{2}+20x+6=0$$

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.

$$x=\frac{-20±\sqrt{20^{2}-4\times 2\times 6}}{2\times 2}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\times 2\times 6}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-20±\sqrt{400-8\times 6}}{2\times 2}$$

Multiply -8 times 6.

$$x=\frac{-20±\sqrt{400-48}}{2\times 2}$$

Add 400 to -48.

$$x=\frac{-20±\sqrt{352}}{2\times 2}$$

Take the square root of 352.

$$x=\frac{-20±4\sqrt{22}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-20±4\sqrt{22}}{4}$$

Now solve the equation x=\frac{-20±4\sqrt{22}}{4} when ± is plus. Add -20 to 4\sqrt{22}.

$$x=\frac{4\sqrt{22}-20}{4}$$

Divide -20+4\sqrt{22} by 4.

$$x=\sqrt{22}-5$$

Now solve the equation x=\frac{-20±4\sqrt{22}}{4} when ± is minus. Subtract 4\sqrt{22} from -20.

$$x=\frac{-4\sqrt{22}-20}{4}$$

Divide -20-4\sqrt{22} by 4.

$$x=-\sqrt{22}-5$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -5+\sqrt{22} for x_{1} and -5-\sqrt{22} for x_{2}.

$$2x^{2}+20x+6=2\left(x-\left(\sqrt{22}-5\right)\right)\left(x-\left(-\sqrt{22}-5\right)\right)$$

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

$$x ^ 2 +10x +3 = 0$$

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

$$r + s = -10 rs = 3$$

Two numbers r and s sum up to -10 exactly when the average of the two numbers is \frac{1}{2}*-10 = -5. 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>

$$r = -5 - u s = -5 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 3

$$(-5 - u) (-5 + u) = 3$$

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

$$25 - u^2 = 3$$

Simplify the expression by subtracting 25 on both sides

$$-u^2 = 3-25 = -22$$

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

$$u^2 = 22 u = \pm\sqrt{22} = \pm \sqrt{22}$$

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

$$r =-5 - \sqrt{22} = -9.690 s = -5 + \sqrt{22} = -0.310$$

Evaluate

$$2\left(x^{2}+10x+3\right)$$