Solve 3x^2+6x-12

$$3 x ^ { 2 } + 6 x - 12$$

Factor

$$3\left(x-\left(-\sqrt{5}-1\right)\right)\left(x-\left(\sqrt{5}-1\right)\right)$$

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.

$$3x^{2}+6x-12=0$$

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.

$$x=\frac{-6±\sqrt{6^{2}-4\times 3\left(-12\right)}}{2\times 3}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 3\left(-12\right)}}{2\times 3}$$

Multiply -4 times 3.

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

Multiply -12 times -12.

$$x=\frac{-6±\sqrt{36+144}}{2\times 3}$$

Add 36 to 144.

$$x=\frac{-6±\sqrt{180}}{2\times 3}$$

Take the square root of 180.

$$x=\frac{-6±6\sqrt{5}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-6±6\sqrt{5}}{6}$$

Now solve the equation x=\frac{-6±6\sqrt{5}}{6} when ± is plus. Add -6 to 6\sqrt{5}.

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

Divide -6+6\sqrt{5} by 6.

$$x=\sqrt{5}-1$$

Now solve the equation x=\frac{-6±6\sqrt{5}}{6} when ± is minus. Subtract 6\sqrt{5} from -6.

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

Divide -6-6\sqrt{5} by 6.

$$x=-\sqrt{5}-1$$

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

$$3x^{2}+6x-12=3\left(x-\left(\sqrt{5}-1\right)\right)\left(x-\left(-\sqrt{5}-1\right)\right)$$

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

$$x ^ 2 +2x -4 = 0$$

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

$$r + s = -2 rs = -4$$

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

$$r = -1 - u s = -1 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -4

$$(-1 - u) (-1 + u) = -4$$

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

$$1 - u^2 = -4$$

Simplify the expression by subtracting 1 on both sides

$$-u^2 = -4-1 = -5$$

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

$$u^2 = 5 u = \pm\sqrt{5} = \pm \sqrt{5}$$

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

$$r =-1 - \sqrt{5} = -3.236 s = -1 + \sqrt{5} = 1.236$$

Evaluate

$$3\left(x^{2}+2x-4\right)$$