Solve x^2+10x+5

$$x ^ { 2 } + 10 x + 5$$

Factor

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

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.

$$x^{2}+10x+5=0$$

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.

$$x=\frac{-10±\sqrt{10^{2}-4\times 5}}{2}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\times 5}}{2}$$

Multiply -4 times 5.

$$x=\frac{-10±\sqrt{100-20}}{2}$$

Add 100 to -20.

$$x=\frac{-10±\sqrt{80}}{2}$$

Take the square root of 80.

$$x=\frac{-10±4\sqrt{5}}{2}$$

Now solve the equation x=\frac{-10±4\sqrt{5}}{2} when ± is plus. Add -10 to 4\sqrt{5}.

$$x=\frac{4\sqrt{5}-10}{2}$$

Divide -10+4\sqrt{5} by 2.

$$x=2\sqrt{5}-5$$

Now solve the equation x=\frac{-10±4\sqrt{5}}{2} when ± is minus. Subtract 4\sqrt{5} from -10.

$$x=\frac{-4\sqrt{5}-10}{2}$$

Divide -10-4\sqrt{5} by 2.

$$x=-2\sqrt{5}-5$$

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

$$x^{2}+10x+5=\left(x-\left(2\sqrt{5}-5\right)\right)\left(x-\left(-2\sqrt{5}-5\right)\right)$$

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.

$$x ^ 2 +10x +5 = 0$$

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

$$r + s = -10 rs = 5$$

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 = 5

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

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

$$25 - u^2 = 5$$

Simplify the expression by subtracting 25 on both sides

$$-u^2 = 5-25 = -20$$

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

$$u^2 = 20 u = \pm\sqrt{20} = \pm \sqrt{20}$$

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

$$r =-5 - \sqrt{20} = -9.472 s = -5 + \sqrt{20} = -0.528$$

Evaluate

$$x^{2}+10x+5$$