Solve -2x^2+12x-6

$$- 2 x ^ { 2 } + 12 x - 6$$

Factor

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

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.

$$-2x^{2}+12x-6=0$$

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.

$$x=\frac{-12±\sqrt{12^{2}-4\left(-2\right)\left(-6\right)}}{2\left(-2\right)}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\left(-2\right)\left(-6\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-12±\sqrt{144+8\left(-6\right)}}{2\left(-2\right)}$$

Multiply 8 times -6.

$$x=\frac{-12±\sqrt{144-48}}{2\left(-2\right)}$$

Add 144 to -48.

$$x=\frac{-12±\sqrt{96}}{2\left(-2\right)}$$

Take the square root of 96.

$$x=\frac{-12±4\sqrt{6}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-12±4\sqrt{6}}{-4}$$

Now solve the equation x=\frac{-12±4\sqrt{6}}{-4} when ± is plus. Add -12 to 4\sqrt{6}.

$$x=\frac{4\sqrt{6}-12}{-4}$$

Divide -12+4\sqrt{6} by -4.

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

Now solve the equation x=\frac{-12±4\sqrt{6}}{-4} when ± is minus. Subtract 4\sqrt{6} from -12.

$$x=\frac{-4\sqrt{6}-12}{-4}$$

Divide -12-4\sqrt{6} by -4.

$$x=\sqrt{6}+3$$

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

$$-2x^{2}+12x-6=-2\left(x-\left(3-\sqrt{6}\right)\right)\left(x-\left(\sqrt{6}+3\right)\right)$$

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.

$$x ^ 2 -6x +3 = 0$$

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

$$r + s = 6 rs = 3$$

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 = 3

$$(3 - u) (3 + u) = 3$$

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

$$9 - u^2 = 3$$

Simplify the expression by subtracting 9 on both sides

$$-u^2 = 3-9 = -6$$

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

$$u^2 = 6 u = \pm\sqrt{6} = \pm \sqrt{6}$$

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

$$r =3 - \sqrt{6} = 0.551 s = 3 + \sqrt{6} = 5.449$$

Evaluate

$$-2x^{2}+12x-6$$