Solve 6x^2+36x+38

$$6 x ^ { 2 } + 36 x + 38$$

Factor

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

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.

$$6x^{2}+36x+38=0$$

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.

$$x=\frac{-36±\sqrt{36^{2}-4\times 6\times 38}}{2\times 6}$$

Square 36.

$$x=\frac{-36±\sqrt{1296-4\times 6\times 38}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-36±\sqrt{1296-24\times 38}}{2\times 6}$$

Multiply -24 times 38.

$$x=\frac{-36±\sqrt{1296-912}}{2\times 6}$$

Add 1296 to -912.

$$x=\frac{-36±\sqrt{384}}{2\times 6}$$

Take the square root of 384.

$$x=\frac{-36±8\sqrt{6}}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{-36±8\sqrt{6}}{12}$$

Now solve the equation x=\frac{-36±8\sqrt{6}}{12} when ± is plus. Add -36 to 8\sqrt{6}.

$$x=\frac{8\sqrt{6}-36}{12}$$

Divide -36+8\sqrt{6} by 12.

$$x=\frac{2\sqrt{6}}{3}-3$$

Now solve the equation x=\frac{-36±8\sqrt{6}}{12} when ± is minus. Subtract 8\sqrt{6} from -36.

$$x=\frac{-8\sqrt{6}-36}{12}$$

Divide -36-8\sqrt{6} by 12.

$$x=-\frac{2\sqrt{6}}{3}-3$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -3+\frac{2\sqrt{6}}{3} for x_{1} and -3-\frac{2\sqrt{6}}{3} for x_{2}.

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

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 6

$$x ^ 2 +6x +\frac{19}{3} = 0$$

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

$$r + s = -6 rs = \frac{19}{3}$$

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

$$r = -3 - u s = -3 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{19}{3}

$$(-3 - u) (-3 + u) = \frac{19}{3}$$

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

$$9 - u^2 = \frac{19}{3}$$

Simplify the expression by subtracting 9 on both sides

$$-u^2 = \frac{19}{3}-9 = -\frac{8}{3}$$

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

$$u^2 = \frac{8}{3} u = \pm\sqrt{\frac{8}{3}} = \pm \frac{\sqrt{8}}{\sqrt{3}}$$

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

$$r =-3 - \frac{\sqrt{8}}{\sqrt{3}} = -4.633 s = -3 + \frac{\sqrt{8}}{\sqrt{3}} = -1.367$$

Evaluate

$$6x^{2}+36x+38$$