Solve 3x^2-36x-6

$$3 x ^ { 2 } - 36 x - 6$$

Factor

$$3\left(x-\left(6-\sqrt{38}\right)\right)\left(x-\left(\sqrt{38}+6\right)\right)$$

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.

$$3x^{2}-36x-6=0$$

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.

$$x=\frac{-\left(-36\right)±\sqrt{\left(-36\right)^{2}-4\times 3\left(-6\right)}}{2\times 3}$$

Square -36.

$$x=\frac{-\left(-36\right)±\sqrt{1296-4\times 3\left(-6\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-36\right)±\sqrt{1296-12\left(-6\right)}}{2\times 3}$$

Multiply -12 times -6.

$$x=\frac{-\left(-36\right)±\sqrt{1296+72}}{2\times 3}$$

Add 1296 to 72.

$$x=\frac{-\left(-36\right)±\sqrt{1368}}{2\times 3}$$

Take the square root of 1368.

$$x=\frac{-\left(-36\right)±6\sqrt{38}}{2\times 3}$$

The opposite of -36 is 36.

$$x=\frac{36±6\sqrt{38}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{36±6\sqrt{38}}{6}$$

Now solve the equation x=\frac{36±6\sqrt{38}}{6} when ± is plus. Add 36 to 6\sqrt{38}.

$$x=\frac{6\sqrt{38}+36}{6}$$

Divide 36+6\sqrt{38} by 6.

$$x=\sqrt{38}+6$$

Now solve the equation x=\frac{36±6\sqrt{38}}{6} when ± is minus. Subtract 6\sqrt{38} from 36.

$$x=\frac{36-6\sqrt{38}}{6}$$

Divide 36-6\sqrt{38} by 6.

$$x=6-\sqrt{38}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 6+\sqrt{38} for x_{1} and 6-\sqrt{38} for x_{2}.

$$3x^{2}-36x-6=3\left(x-\left(\sqrt{38}+6\right)\right)\left(x-\left(6-\sqrt{38}\right)\right)$$

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 3

$$x ^ 2 -12x -2 = 0$$

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

$$r + s = 12 rs = -2$$

Two numbers r and s sum up to 12 exactly when the average of the two numbers is \frac{1}{2}*12 = 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>

$$r = 6 - u s = 6 + u$$

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

$$(6 - u) (6 + u) = -2$$

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

$$36 - u^2 = -2$$

Simplify the expression by subtracting 36 on both sides

$$-u^2 = -2-36 = -38$$

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

$$u^2 = 38 u = \pm\sqrt{38} = \pm \sqrt{38}$$

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

$$r =6 - \sqrt{38} = -0.164 s = 6 + \sqrt{38} = 12.164$$

Evaluate

$$3\left(x^{2}-12x-2\right)$$