Solve 2y^2-10y-3

$$2 y ^ { 2 } - 10 y - 3$$

Factor

$$2\left(y-\frac{5-\sqrt{31}}{2}\right)\left(y-\frac{\sqrt{31}+5}{2}\right)$$

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.

$$2y^{2}-10y-3=0$$

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.

$$y=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 2\left(-3\right)}}{2\times 2}$$

Square -10.

$$y=\frac{-\left(-10\right)±\sqrt{100-4\times 2\left(-3\right)}}{2\times 2}$$

Multiply -4 times 2.

$$y=\frac{-\left(-10\right)±\sqrt{100-8\left(-3\right)}}{2\times 2}$$

Multiply -8 times -3.

$$y=\frac{-\left(-10\right)±\sqrt{100+24}}{2\times 2}$$

Add 100 to 24.

$$y=\frac{-\left(-10\right)±\sqrt{124}}{2\times 2}$$

Take the square root of 124.

$$y=\frac{-\left(-10\right)±2\sqrt{31}}{2\times 2}$$

The opposite of -10 is 10.

$$y=\frac{10±2\sqrt{31}}{2\times 2}$$

Multiply 2 times 2.

$$y=\frac{10±2\sqrt{31}}{4}$$

Now solve the equation y=\frac{10±2\sqrt{31}}{4} when ± is plus. Add 10 to 2\sqrt{31}.

$$y=\frac{2\sqrt{31}+10}{4}$$

Divide 10+2\sqrt{31} by 4.

$$y=\frac{\sqrt{31}+5}{2}$$

Now solve the equation y=\frac{10±2\sqrt{31}}{4} when ± is minus. Subtract 2\sqrt{31} from 10.

$$y=\frac{10-2\sqrt{31}}{4}$$

Divide 10-2\sqrt{31} by 4.

$$y=\frac{5-\sqrt{31}}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5+\sqrt{31}}{2} for x_{1} and \frac{5-\sqrt{31}}{2} for x_{2}.

$$2y^{2}-10y-3=2\left(y-\frac{\sqrt{31}+5}{2}\right)\left(y-\frac{5-\sqrt{31}}{2}\right)$$

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 2

$$x ^ 2 -5x -\frac{3}{2} = 0$$

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

$$r + s = 5 rs = -\frac{3}{2}$$

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

$$r = \frac{5}{2} - u s = \frac{5}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{2}

$$(\frac{5}{2} - u) (\frac{5}{2} + u) = -\frac{3}{2}$$

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

$$\frac{25}{4} - u^2 = -\frac{3}{2}$$

Simplify the expression by subtracting \frac{25}{4} on both sides

$$-u^2 = -\frac{3}{2}-\frac{25}{4} = -\frac{31}{4}$$

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

$$u^2 = \frac{31}{4} u = \pm\sqrt{\frac{31}{4}} = \pm \frac{\sqrt{31}}{2}$$

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

$$r =\frac{5}{2} - \frac{\sqrt{31}}{2} = -0.284 s = \frac{5}{2} + \frac{\sqrt{31}}{2} = 5.284$$

Evaluate

$$2y^{2}-10y-3$$