Solve x^2-10x-320

$$x ^ { 2 } - 10 x - 320$$

Factor

$$\left(x-\left(5-\sqrt{345}\right)\right)\left(x-\left(\sqrt{345}+5\right)\right)$$

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.

$$x^{2}-10x-320=0$$

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.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-320\right)}}{2}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-320\right)}}{2}$$

Multiply -4 times -320.

$$x=\frac{-\left(-10\right)±\sqrt{100+1280}}{2}$$

Add 100 to 1280.

$$x=\frac{-\left(-10\right)±\sqrt{1380}}{2}$$

Take the square root of 1380.

$$x=\frac{-\left(-10\right)±2\sqrt{345}}{2}$$

The opposite of -10 is 10.

$$x=\frac{10±2\sqrt{345}}{2}$$

Now solve the equation x=\frac{10±2\sqrt{345}}{2} when ± is plus. Add 10 to 2\sqrt{345}.

$$x=\frac{2\sqrt{345}+10}{2}$$

Divide 10+2\sqrt{345} by 2.

$$x=\sqrt{345}+5$$

Now solve the equation x=\frac{10±2\sqrt{345}}{2} when ± is minus. Subtract 2\sqrt{345} from 10.

$$x=\frac{10-2\sqrt{345}}{2}$$

Divide 10-2\sqrt{345} by 2.

$$x=5-\sqrt{345}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 5+\sqrt{345} for x_{1} and 5-\sqrt{345} for x_{2}.

$$x^{2}-10x-320=\left(x-\left(\sqrt{345}+5\right)\right)\left(x-\left(5-\sqrt{345}\right)\right)$$

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.

$$x ^ 2 -10x -320 = 0$$

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

$$r + s = 10 rs = -320$$

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 = -320

$$(5 - u) (5 + u) = -320$$

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

$$25 - u^2 = -320$$

Simplify the expression by subtracting 25 on both sides

$$-u^2 = -320-25 = -345$$

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

$$u^2 = 345 u = \pm\sqrt{345} = \pm \sqrt{345}$$

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

$$r =5 - \sqrt{345} = -13.574 s = 5 + \sqrt{345} = 23.574$$

Evaluate

$$x^{2}-10x-320$$