Solve 6u^2+24u-36

$$6 u ^ { 2 } + 24 u - 36$$

Factor

$$6\left(u-\left(-\sqrt{10}-2\right)\right)\left(u-\left(\sqrt{10}-2\right)\right)$$

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.

$$6u^{2}+24u-36=0$$

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.

$$u=\frac{-24±\sqrt{24^{2}-4\times 6\left(-36\right)}}{2\times 6}$$

Square 24.

$$u=\frac{-24±\sqrt{576-4\times 6\left(-36\right)}}{2\times 6}$$

Multiply -4 times 6.

$$u=\frac{-24±\sqrt{576-24\left(-36\right)}}{2\times 6}$$

Multiply -24 times -36.

$$u=\frac{-24±\sqrt{576+864}}{2\times 6}$$

Add 576 to 864.

$$u=\frac{-24±\sqrt{1440}}{2\times 6}$$

Take the square root of 1440.

$$u=\frac{-24±12\sqrt{10}}{2\times 6}$$

Multiply 2 times 6.

$$u=\frac{-24±12\sqrt{10}}{12}$$

Now solve the equation u=\frac{-24±12\sqrt{10}}{12} when ± is plus. Add -24 to 12\sqrt{10}.

$$u=\frac{12\sqrt{10}-24}{12}$$

Divide -24+12\sqrt{10} by 12.

$$u=\sqrt{10}-2$$

Now solve the equation u=\frac{-24±12\sqrt{10}}{12} when ± is minus. Subtract 12\sqrt{10} from -24.

$$u=\frac{-12\sqrt{10}-24}{12}$$

Divide -24-12\sqrt{10} by 12.

$$u=-\sqrt{10}-2$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -2+\sqrt{10} for x_{1} and -2-\sqrt{10} for x_{2}.

$$6u^{2}+24u-36=6\left(u-\left(\sqrt{10}-2\right)\right)\left(u-\left(-\sqrt{10}-2\right)\right)$$

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

$$x ^ 2 +4x -6 = 0$$

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

$$r + s = -4 rs = -6$$

Two numbers r and s sum up to -4 exactly when the average of the two numbers is \frac{1}{2}*-4 = -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>

$$r = -2 - u s = -2 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -6

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

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

$$4 - u^2 = -6$$

Simplify the expression by subtracting 4 on both sides

$$-u^2 = -6-4 = -10$$

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

$$u^2 = 10 u = \pm\sqrt{10} = \pm \sqrt{10}$$

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

$$r =-2 - \sqrt{10} = -5.162 s = -2 + \sqrt{10} = 1.162$$

Evaluate

$$6\left(u^{2}+4u-6\right)$$