Solve x^2-21x-19350

$$x ^ { 2 } - 21 x - 19350$$

Factor

$$\left(x-150\right)\left(x+129\right)$$

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.

$$x^{2}-21x-19350=0$$

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.

$$x=\frac{-\left(-21\right)±\sqrt{\left(-21\right)^{2}-4\left(-19350\right)}}{2}$$

Square -21.

$$x=\frac{-\left(-21\right)±\sqrt{441-4\left(-19350\right)}}{2}$$

Multiply -4 times -19350.

$$x=\frac{-\left(-21\right)±\sqrt{441+77400}}{2}$$

Add 441 to 77400.

$$x=\frac{-\left(-21\right)±\sqrt{77841}}{2}$$

Take the square root of 77841.

$$x=\frac{-\left(-21\right)±279}{2}$$

The opposite of -21 is 21.

$$x=\frac{21±279}{2}$$

Now solve the equation x=\frac{21±279}{2} when ± is plus. Add 21 to 279.

$$x=\frac{300}{2}$$

Divide 300 by 2.

$$x=150$$

Now solve the equation x=\frac{21±279}{2} when ± is minus. Subtract 279 from 21.

$$x=-\frac{258}{2}$$

Divide -258 by 2.

$$x=-129$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 150 for x_{1} and -129 for x_{2}.

$$x^{2}-21x-19350=\left(x-150\right)\left(x-\left(-129\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$x^{2}-21x-19350=\left(x-150\right)\left(x+129\right)$$

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.

$$x ^ 2 -21x -19350 = 0$$

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

$$r + s = 21 rs = -19350$$

Two numbers r and s sum up to 21 exactly when the average of the two numbers is \frac{1}{2}*21 = \frac{21}{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.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{21}{2} - u s = \frac{21}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -19350

$$(\frac{21}{2} - u) (\frac{21}{2} + u) = -19350$$

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

$$\frac{441}{4} - u^2 = -19350$$

Simplify the expression by subtracting \frac{441}{4} on both sides

$$-u^2 = -19350-\frac{441}{4} = -\frac{77841}{4}$$

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

$$u^2 = \frac{77841}{4} u = \pm\sqrt{\frac{77841}{4}} = \pm \frac{279}{2}$$

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

$$r =\frac{21}{2} - \frac{279}{2} = -129 s = \frac{21}{2} + \frac{279}{2} = 150$$

Evaluate

$$\left(x-150\right)\left(x+129\right)$$