Solve 6y^2+22y+5

$$6 y ^ { 2 } + 22 y + 5$$

Factor

$$6\left(y-\frac{-\sqrt{91}-11}{6}\right)\left(y-\frac{\sqrt{91}-11}{6}\right)$$

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.

$$6y^{2}+22y+5=0$$

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.

$$y=\frac{-22±\sqrt{22^{2}-4\times 6\times 5}}{2\times 6}$$

Square 22.

$$y=\frac{-22±\sqrt{484-4\times 6\times 5}}{2\times 6}$$

Multiply -4 times 6.

$$y=\frac{-22±\sqrt{484-24\times 5}}{2\times 6}$$

Multiply -24 times 5.

$$y=\frac{-22±\sqrt{484-120}}{2\times 6}$$

Add 484 to -120.

$$y=\frac{-22±\sqrt{364}}{2\times 6}$$

Take the square root of 364.

$$y=\frac{-22±2\sqrt{91}}{2\times 6}$$

Multiply 2 times 6.

$$y=\frac{-22±2\sqrt{91}}{12}$$

Now solve the equation y=\frac{-22±2\sqrt{91}}{12} when ± is plus. Add -22 to 2\sqrt{91}.

$$y=\frac{2\sqrt{91}-22}{12}$$

Divide -22+2\sqrt{91} by 12.

$$y=\frac{\sqrt{91}-11}{6}$$

Now solve the equation y=\frac{-22±2\sqrt{91}}{12} when ± is minus. Subtract 2\sqrt{91} from -22.

$$y=\frac{-2\sqrt{91}-22}{12}$$

Divide -22-2\sqrt{91} by 12.

$$y=\frac{-\sqrt{91}-11}{6}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-11+\sqrt{91}}{6} for x_{1} and \frac{-11-\sqrt{91}}{6} for x_{2}.

$$6y^{2}+22y+5=6\left(y-\frac{\sqrt{91}-11}{6}\right)\left(y-\frac{-\sqrt{91}-11}{6}\right)$$

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

$$x ^ 2 +\frac{11}{3}x +\frac{5}{6} = 0$$

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

$$r + s = -\frac{11}{3} rs = \frac{5}{6}$$

Two numbers r and s sum up to -\frac{11}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{11}{3} = -\frac{11}{6}. 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>

$$r = -\frac{11}{6} - u s = -\frac{11}{6} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{5}{6}

$$(-\frac{11}{6} - u) (-\frac{11}{6} + u) = \frac{5}{6}$$

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

$$\frac{121}{36} - u^2 = \frac{5}{6}$$

Simplify the expression by subtracting \frac{121}{36} on both sides

$$-u^2 = \frac{5}{6}-\frac{121}{36} = -\frac{91}{36}$$

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

$$u^2 = \frac{91}{36} u = \pm\sqrt{\frac{91}{36}} = \pm \frac{\sqrt{91}}{6}$$

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

$$r =-\frac{11}{6} - \frac{\sqrt{91}}{6} = -3.423 s = -\frac{11}{6} + \frac{\sqrt{91}}{6} = -0.243$$

Evaluate

$$6y^{2}+22y+5$$