Solve 10q^2+3q+45

$$10 q ^ { 2 } + 3 q + 45$$

Evaluate

$$10q^{2}+3q+45$$

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 10

$$x ^ 2 +\frac{3}{10}x +\frac{9}{2} = 0$$

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

$$r + s = -\frac{3}{10} rs = \frac{9}{2}$$

Two numbers r and s sum up to -\frac{3}{10} exactly when the average of the two numbers is \frac{1}{2}*-\frac{3}{10} = -\frac{3}{20}. 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>

$$r = -\frac{3}{20} - u s = -\frac{3}{20} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{9}{2}

$$(-\frac{3}{20} - u) (-\frac{3}{20} + u) = \frac{9}{2}$$

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

$$\frac{9}{400} - u^2 = \frac{9}{2}$$

Simplify the expression by subtracting \frac{9}{400} on both sides

$$-u^2 = \frac{9}{2}-\frac{9}{400} = \frac{1791}{400}$$

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

$$u^2 = -\frac{1791}{400} u = \pm\sqrt{-\frac{1791}{400}} = \pm \frac{\sqrt{1791}}{20}i$$

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

$$r =-\frac{3}{20} - \frac{\sqrt{1791}}{20}i = -0.150 - 2.116i s = -\frac{3}{20} + \frac{\sqrt{1791}}{20}i = -0.150 + 2.116i$$

Differentiate w.r.t. q

$$20q+3$$

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}.

$$2\times 10q^{2-1}+3q^{1-1}$$

Multiply 2 times 10.

$$20q^{2-1}+3q^{1-1}$$

Subtract 1 from 2.

$$20q^{1}+3q^{1-1}$$

Subtract 1 from 1.

$$20q^{1}+3q^{0}$$

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

$$20q+3q^{0}$$

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

$$20q+3\times 1$$

For any term t, t\times 1=t and 1t=t.

$$20q+3$$