Solve 6x^2+18x+3

$$6 x ^ { 2 } + 18 x + 3$$

Factor

$$6\left(x-\frac{-\sqrt{7}-3}{2}\right)\left(x-\frac{\sqrt{7}-3}{2}\right)$$

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.

$$6x^{2}+18x+3=0$$

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.

$$x=\frac{-18±\sqrt{18^{2}-4\times 6\times 3}}{2\times 6}$$

Square 18.

$$x=\frac{-18±\sqrt{324-4\times 6\times 3}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-18±\sqrt{324-24\times 3}}{2\times 6}$$

Multiply -24 times 3.

$$x=\frac{-18±\sqrt{324-72}}{2\times 6}$$

Add 324 to -72.

$$x=\frac{-18±\sqrt{252}}{2\times 6}$$

Take the square root of 252.

$$x=\frac{-18±6\sqrt{7}}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{-18±6\sqrt{7}}{12}$$

Now solve the equation x=\frac{-18±6\sqrt{7}}{12} when ± is plus. Add -18 to 6\sqrt{7}.

$$x=\frac{6\sqrt{7}-18}{12}$$

Divide -18+6\sqrt{7} by 12.

$$x=\frac{\sqrt{7}-3}{2}$$

Now solve the equation x=\frac{-18±6\sqrt{7}}{12} when ± is minus. Subtract 6\sqrt{7} from -18.

$$x=\frac{-6\sqrt{7}-18}{12}$$

Divide -18-6\sqrt{7} by 12.

$$x=\frac{-\sqrt{7}-3}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-3+\sqrt{7}}{2} for x_{1} and \frac{-3-\sqrt{7}}{2} for x_{2}.

$$6x^{2}+18x+3=6\left(x-\frac{\sqrt{7}-3}{2}\right)\left(x-\frac{-\sqrt{7}-3}{2}\right)$$

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 6

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

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

$$r + s = -3 rs = \frac{1}{2}$$

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

$$r = -\frac{3}{2} - u s = -\frac{3}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{2}

$$(-\frac{3}{2} - u) (-\frac{3}{2} + u) = \frac{1}{2}$$

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

$$\frac{9}{4} - u^2 = \frac{1}{2}$$

Simplify the expression by subtracting \frac{9}{4} on both sides

$$-u^2 = \frac{1}{2}-\frac{9}{4} = -\frac{7}{4}$$

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

$$u^2 = \frac{7}{4} u = \pm\sqrt{\frac{7}{4}} = \pm \frac{\sqrt{7}}{2}$$

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

$$r =-\frac{3}{2} - \frac{\sqrt{7}}{2} = -2.823 s = -\frac{3}{2} + \frac{\sqrt{7}}{2} = -0.177$$

Evaluate

$$6x^{2}+18x+3$$