Solve =2x^2-20x+510

$$= 2 x ^ { 2 } - 20 x + 510$$

Factor

$$2\left(x^{2}-10x+255\right)$$

Solution Steps

Factor out 2. Polynomial x^{2}-10x+255 is not factored since it does not have any rational roots.

$$2\left(x^{2}-10x+255\right)$$

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.

$$2x^{2}-20x+510=0$$

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.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 2\times 510}}{2\times 2}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 2\times 510}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-20\right)±\sqrt{400-8\times 510}}{2\times 2}$$

Multiply -8 times 510.

$$x=\frac{-\left(-20\right)±\sqrt{400-4080}}{2\times 2}$$

Add 400 to -4080.

$$x=\frac{-\left(-20\right)±\sqrt{-3680}}{2\times 2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions. Quadratic polynomial cannot be factored.

$$2x^{2}-20x+510$$

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.This is achieved by dividing both sides of the equation by 2

$$x ^ 2 -10x +255 = 0$$

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

$$r + s = 10 rs = 255$$

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 = 255

$$(5 - u) (5 + u) = 255$$

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

$$25 - u^2 = 255$$

Simplify the expression by subtracting 25 on both sides

$$-u^2 = 255-25 = 230$$

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

$$u^2 = -230 u = \pm\sqrt{-230} = \pm \sqrt{230}i$$

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

$$r =5 - \sqrt{230}i s = 5 + \sqrt{230}i$$

Evaluate

$$2\left(x^{2}-10x+255\right)$$