Solve {l}{x^2+4x}{-300}

$$\left. \begin{array} { l } { x ^ { 2 } + 4 x } \\ { - 300 } \end{array} \right.$$

Factor

$$\left(x-\left(-4\sqrt{19}-2\right)\right)\left(x-\left(4\sqrt{19}-2\right)\right)$$

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.

$$x^{2}+4x-300=0$$

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.

$$x=\frac{-4±\sqrt{4^{2}-4\left(-300\right)}}{2}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\left(-300\right)}}{2}$$

Multiply -4 times -300.

$$x=\frac{-4±\sqrt{16+1200}}{2}$$

Add 16 to 1200.

$$x=\frac{-4±\sqrt{1216}}{2}$$

Take the square root of 1216.

$$x=\frac{-4±8\sqrt{19}}{2}$$

Now solve the equation x=\frac{-4±8\sqrt{19}}{2} when ± is plus. Add -4 to 8\sqrt{19}.

$$x=\frac{8\sqrt{19}-4}{2}$$

Divide -4+8\sqrt{19} by 2.

$$x=4\sqrt{19}-2$$

Now solve the equation x=\frac{-4±8\sqrt{19}}{2} when ± is minus. Subtract 8\sqrt{19} from -4.

$$x=\frac{-8\sqrt{19}-4}{2}$$

Divide -4-8\sqrt{19} by 2.

$$x=-4\sqrt{19}-2$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -2+4\sqrt{19} for x_{1} and -2-4\sqrt{19} for x_{2}.

$$x^{2}+4x-300=\left(x-\left(4\sqrt{19}-2\right)\right)\left(x-\left(-4\sqrt{19}-2\right)\right)$$

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.

$$x ^ 2 +4x -300 = 0$$

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

$$r + s = -4 rs = -300$$

Two numbers r and s sum up to -4 exactly when the average of the two numbers is \frac{1}{2}*-4 = -2. 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-gzdabgg4ehffg0hf.b01.azurefd.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = -2 - u s = -2 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -300

$$(-2 - u) (-2 + u) = -300$$

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

$$4 - u^2 = -300$$

Simplify the expression by subtracting 4 on both sides

$$-u^2 = -300-4 = -304$$

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

$$u^2 = 304 u = \pm\sqrt{304} = \pm \sqrt{304}$$

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

$$r =-2 - \sqrt{304} = -19.436 s = -2 + \sqrt{304} = 15.436$$

Evaluate

$$x^{2}+4x-300$$