Solve {l}{5x^2-2x}{+10}

$$\left. \begin{array} { l } { 5 x ^ { 2 } - 2 x } \\ { + 10 } \end{array} \right.$$

Evaluate

$$5x^{2}-2x+10$$

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 5

$$x ^ 2 -\frac{2}{5}x +2 = 0$$

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

$$r + s = \frac{2}{5} rs = 2$$

Two numbers r and s sum up to \frac{2}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{2}{5} = \frac{1}{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>

$$r = \frac{1}{5} - u s = \frac{1}{5} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 2

$$(\frac{1}{5} - u) (\frac{1}{5} + u) = 2$$

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

$$\frac{1}{25} - u^2 = 2$$

Simplify the expression by subtracting \frac{1}{25} on both sides

$$-u^2 = 2-\frac{1}{25} = \frac{49}{25}$$

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

$$u^2 = -\frac{49}{25} u = \pm\sqrt{-\frac{49}{25}} = \pm \frac{7}{5}i$$

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

$$r =\frac{1}{5} - \frac{7}{5}i = 0.200 - 1.400i s = \frac{1}{5} + \frac{7}{5}i = 0.200 + 1.400i$$

Differentiate w.r.t. x

$$10x-2$$

Steps Using Derivative Rule for Sum

The derivative of a polynomial is the sum of the derivatives of its terms. The derivative of a constant term is 0. The derivative of ax^{n} is nax^{n-1}.

$$2\times 5x^{2-1}-2x^{1-1}$$

Multiply 2 times 5.

$$10x^{2-1}-2x^{1-1}$$

Subtract 1 from 2.

$$10x^{1}-2x^{1-1}$$

Subtract 1 from 1.

$$10x^{1}-2x^{0}$$

For any term t, t^{1}=t.

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

For any term t except 0, t^{0}=1.

$$10x-2$$