Solve x^2+28x+5

$$x ^ { 2 } + 28 x + 5$$

Factor

$$\left(x-\left(-\sqrt{191}-14\right)\right)\left(x-\left(\sqrt{191}-14\right)\right)$$

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.

$$x^{2}+28x+5=0$$

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.

$$x=\frac{-28±\sqrt{28^{2}-4\times 5}}{2}$$

Square 28.

$$x=\frac{-28±\sqrt{784-4\times 5}}{2}$$

Multiply -4 times 5.

$$x=\frac{-28±\sqrt{784-20}}{2}$$

Add 784 to -20.

$$x=\frac{-28±\sqrt{764}}{2}$$

Take the square root of 764.

$$x=\frac{-28±2\sqrt{191}}{2}$$

Now solve the equation x=\frac{-28±2\sqrt{191}}{2} when ± is plus. Add -28 to 2\sqrt{191}.

$$x=\frac{2\sqrt{191}-28}{2}$$

Divide -28+2\sqrt{191} by 2.

$$x=\sqrt{191}-14$$

Now solve the equation x=\frac{-28±2\sqrt{191}}{2} when ± is minus. Subtract 2\sqrt{191} from -28.

$$x=\frac{-2\sqrt{191}-28}{2}$$

Divide -28-2\sqrt{191} by 2.

$$x=-\sqrt{191}-14$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -14+\sqrt{191} for x_{1} and -14-\sqrt{191} for x_{2}.

$$x^{2}+28x+5=\left(x-\left(\sqrt{191}-14\right)\right)\left(x-\left(-\sqrt{191}-14\right)\right)$$

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.

$$x ^ 2 +28x +5 = 0$$

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

$$r + s = -28 rs = 5$$

Two numbers r and s sum up to -28 exactly when the average of the two numbers is \frac{1}{2}*-28 = -14. 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>

$$r = -14 - u s = -14 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 5

$$(-14 - u) (-14 + u) = 5$$

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

$$196 - u^2 = 5$$

Simplify the expression by subtracting 196 on both sides

$$-u^2 = 5-196 = -191$$

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

$$u^2 = 191 u = \pm\sqrt{191} = \pm \sqrt{191}$$

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

$$r =-14 - \sqrt{191} = -27.820 s = -14 + \sqrt{191} = -0.180$$

Evaluate

$$x^{2}+28x+5$$