Solve 6y^2-20y-6

$$6 y ^ { 2 } - 20 y - 6$$

Factor

$$6\left(y-\frac{5-\sqrt{34}}{3}\right)\left(y-\frac{\sqrt{34}+5}{3}\right)$$

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.

$$6y^{2}-20y-6=0$$

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.

$$y=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 6\left(-6\right)}}{2\times 6}$$

Square -20.

$$y=\frac{-\left(-20\right)±\sqrt{400-4\times 6\left(-6\right)}}{2\times 6}$$

Multiply -4 times 6.

$$y=\frac{-\left(-20\right)±\sqrt{400-24\left(-6\right)}}{2\times 6}$$

Multiply -24 times -6.

$$y=\frac{-\left(-20\right)±\sqrt{400+144}}{2\times 6}$$

Add 400 to 144.

$$y=\frac{-\left(-20\right)±\sqrt{544}}{2\times 6}$$

Take the square root of 544.

$$y=\frac{-\left(-20\right)±4\sqrt{34}}{2\times 6}$$

The opposite of -20 is 20.

$$y=\frac{20±4\sqrt{34}}{2\times 6}$$

Multiply 2 times 6.

$$y=\frac{20±4\sqrt{34}}{12}$$

Now solve the equation y=\frac{20±4\sqrt{34}}{12} when ± is plus. Add 20 to 4\sqrt{34}.

$$y=\frac{4\sqrt{34}+20}{12}$$

Divide 20+4\sqrt{34} by 12.

$$y=\frac{\sqrt{34}+5}{3}$$

Now solve the equation y=\frac{20±4\sqrt{34}}{12} when ± is minus. Subtract 4\sqrt{34} from 20.

$$y=\frac{20-4\sqrt{34}}{12}$$

Divide 20-4\sqrt{34} by 12.

$$y=\frac{5-\sqrt{34}}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5+\sqrt{34}}{3} for x_{1} and \frac{5-\sqrt{34}}{3} for x_{2}.

$$6y^{2}-20y-6=6\left(y-\frac{\sqrt{34}+5}{3}\right)\left(y-\frac{5-\sqrt{34}}{3}\right)$$

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

$$x ^ 2 -\frac{10}{3}x -1 = 0$$

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

$$r + s = \frac{10}{3} rs = -1$$

Two numbers r and s sum up to \frac{10}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{10}{3} = \frac{5}{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>

$$r = \frac{5}{3} - u s = \frac{5}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -1

$$(\frac{5}{3} - u) (\frac{5}{3} + u) = -1$$

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

$$\frac{25}{9} - u^2 = -1$$

Simplify the expression by subtracting \frac{25}{9} on both sides

$$-u^2 = -1-\frac{25}{9} = -\frac{34}{9}$$

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

$$u^2 = \frac{34}{9} u = \pm\sqrt{\frac{34}{9}} = \pm \frac{\sqrt{34}}{3}$$

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

$$r =\frac{5}{3} - \frac{\sqrt{34}}{3} = -0.277 s = \frac{5}{3} + \frac{\sqrt{34}}{3} = 3.610$$

Evaluate

$$6y^{2}-20y-6$$