Solve x^2+20x+10=?

$$x ^ { 2 } + 20 x + 10 = ?$$

Factor

$$\left(x-\left(-3\sqrt{10}-10\right)\right)\left(x-\left(3\sqrt{10}-10\right)\right)$$

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.

$$x^{2}+20x+10=0$$

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.

$$x=\frac{-20±\sqrt{20^{2}-4\times 10}}{2}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\times 10}}{2}$$

Multiply -4 times 10.

$$x=\frac{-20±\sqrt{400-40}}{2}$$

Add 400 to -40.

$$x=\frac{-20±\sqrt{360}}{2}$$

Take the square root of 360.

$$x=\frac{-20±6\sqrt{10}}{2}$$

Now solve the equation x=\frac{-20±6\sqrt{10}}{2} when ± is plus. Add -20 to 6\sqrt{10}.

$$x=\frac{6\sqrt{10}-20}{2}$$

Divide -20+6\sqrt{10} by 2.

$$x=3\sqrt{10}-10$$

Now solve the equation x=\frac{-20±6\sqrt{10}}{2} when ± is minus. Subtract 6\sqrt{10} from -20.

$$x=\frac{-6\sqrt{10}-20}{2}$$

Divide -20-6\sqrt{10} by 2.

$$x=-3\sqrt{10}-10$$

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

$$x^{2}+20x+10=\left(x-\left(3\sqrt{10}-10\right)\right)\left(x-\left(-3\sqrt{10}-10\right)\right)$$

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.

$$x ^ 2 +20x +10 = 0$$

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

$$r + s = -20 rs = 10$$

Two numbers r and s sum up to -20 exactly when the average of the two numbers is \frac{1}{2}*-20 = -10. 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>

$$r = -10 - u s = -10 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 10

$$(-10 - u) (-10 + u) = 10$$

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

$$100 - u^2 = 10$$

Simplify the expression by subtracting 100 on both sides

$$-u^2 = 10-100 = -90$$

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

$$u^2 = 90 u = \pm\sqrt{90} = \pm \sqrt{90}$$

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

$$r =-10 - \sqrt{90} = -19.487 s = -10 + \sqrt{90} = -0.513$$

Evaluate

$$x^{2}+20x+10$$