Solve x^2-328x+1280

$$x ^ { 2 } - 328 x + 1280$$

Factor

$$\left(x-\left(164-4\sqrt{1601}\right)\right)\left(x-\left(4\sqrt{1601}+164\right)\right)$$

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.

$$x^{2}-328x+1280=0$$

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.

$$x=\frac{-\left(-328\right)±\sqrt{\left(-328\right)^{2}-4\times 1280}}{2}$$

Square -328.

$$x=\frac{-\left(-328\right)±\sqrt{107584-4\times 1280}}{2}$$

Multiply -4 times 1280.

$$x=\frac{-\left(-328\right)±\sqrt{107584-5120}}{2}$$

Add 107584 to -5120.

$$x=\frac{-\left(-328\right)±\sqrt{102464}}{2}$$

Take the square root of 102464.

$$x=\frac{-\left(-328\right)±8\sqrt{1601}}{2}$$

The opposite of -328 is 328.

$$x=\frac{328±8\sqrt{1601}}{2}$$

Now solve the equation x=\frac{328±8\sqrt{1601}}{2} when ± is plus. Add 328 to 8\sqrt{1601}.

$$x=\frac{8\sqrt{1601}+328}{2}$$

Divide 328+8\sqrt{1601} by 2.

$$x=4\sqrt{1601}+164$$

Now solve the equation x=\frac{328±8\sqrt{1601}}{2} when ± is minus. Subtract 8\sqrt{1601} from 328.

$$x=\frac{328-8\sqrt{1601}}{2}$$

Divide 328-8\sqrt{1601} by 2.

$$x=164-4\sqrt{1601}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 164+4\sqrt{1601} for x_{1} and 164-4\sqrt{1601} for x_{2}.

$$x^{2}-328x+1280=\left(x-\left(4\sqrt{1601}+164\right)\right)\left(x-\left(164-4\sqrt{1601}\right)\right)$$

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.

$$x ^ 2 -328x +1280 = 0$$

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

$$r + s = 328 rs = 1280$$

Two numbers r and s sum up to 328 exactly when the average of the two numbers is \frac{1}{2}*328 = 164. 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>

$$r = 164 - u s = 164 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 1280

$$(164 - u) (164 + u) = 1280$$

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

$$26896 - u^2 = 1280$$

Simplify the expression by subtracting 26896 on both sides

$$-u^2 = 1280-26896 = -25616$$

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

$$u^2 = 25616 u = \pm\sqrt{25616} = \pm \sqrt{25616}$$

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

$$r =164 - \sqrt{25616} = 3.950 s = 164 + \sqrt{25616} = 324.050$$

Evaluate

$$x^{2}-328x+1280$$