Solve f(x)=18x^2+30x+2

$$f ( x ) = 18 x ^ { 2 } + 30 x + 2$$

Factor

$$18\left(x-\frac{-\sqrt{21}-5}{6}\right)\left(x-\frac{\sqrt{21}-5}{6}\right)$$

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.

$$18x^{2}+30x+2=0$$

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.

$$x=\frac{-30±\sqrt{30^{2}-4\times 18\times 2}}{2\times 18}$$

Square 30.

$$x=\frac{-30±\sqrt{900-4\times 18\times 2}}{2\times 18}$$

Multiply -4 times 18.

$$x=\frac{-30±\sqrt{900-72\times 2}}{2\times 18}$$

Multiply -72 times 2.

$$x=\frac{-30±\sqrt{900-144}}{2\times 18}$$

Add 900 to -144.

$$x=\frac{-30±\sqrt{756}}{2\times 18}$$

Take the square root of 756.

$$x=\frac{-30±6\sqrt{21}}{2\times 18}$$

Multiply 2 times 18.

$$x=\frac{-30±6\sqrt{21}}{36}$$

Now solve the equation x=\frac{-30±6\sqrt{21}}{36} when ± is plus. Add -30 to 6\sqrt{21}.

$$x=\frac{6\sqrt{21}-30}{36}$$

Divide -30+6\sqrt{21} by 36.

$$x=\frac{\sqrt{21}-5}{6}$$

Now solve the equation x=\frac{-30±6\sqrt{21}}{36} when ± is minus. Subtract 6\sqrt{21} from -30.

$$x=\frac{-6\sqrt{21}-30}{36}$$

Divide -30-6\sqrt{21} by 36.

$$x=\frac{-\sqrt{21}-5}{6}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-5+\sqrt{21}}{6} for x_{1} and \frac{-5-\sqrt{21}}{6} for x_{2}.

$$18x^{2}+30x+2=18\left(x-\frac{\sqrt{21}-5}{6}\right)\left(x-\frac{-\sqrt{21}-5}{6}\right)$$

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.This is achieved by dividing both sides of the equation by 18

$$x ^ 2 +\frac{5}{3}x +\frac{1}{9} = 0$$

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

$$r + s = -\frac{5}{3} rs = \frac{1}{9}$$

Two numbers r and s sum up to -\frac{5}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{5}{3} = -\frac{5}{6}. 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>

$$r = -\frac{5}{6} - u s = -\frac{5}{6} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{9}

$$(-\frac{5}{6} - u) (-\frac{5}{6} + u) = \frac{1}{9}$$

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

$$\frac{25}{36} - u^2 = \frac{1}{9}$$

Simplify the expression by subtracting \frac{25}{36} on both sides

$$-u^2 = \frac{1}{9}-\frac{25}{36} = -\frac{7}{12}$$

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

$$u^2 = \frac{7}{12} u = \pm\sqrt{\frac{7}{12}} = \pm \frac{\sqrt{7}}{\sqrt{12}}$$

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

$$r =-\frac{5}{6} - \frac{\sqrt{7}}{\sqrt{12}} = -1.597 s = -\frac{5}{6} + \frac{\sqrt{7}}{\sqrt{12}} = -0.070$$

Evaluate

$$18x^{2}+30x+2$$