Solve 2x^2-2x-19

$$2 x ^ { 2 } - 2 x - 19$$

Factor

$$2\left(x-\frac{1-\sqrt{39}}{2}\right)\left(x-\frac{\sqrt{39}+1}{2}\right)$$

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.

$$2x^{2}-2x-19=0$$

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.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 2\left(-19\right)}}{2\times 2}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\times 2\left(-19\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-2\right)±\sqrt{4-8\left(-19\right)}}{2\times 2}$$

Multiply -8 times -19.

$$x=\frac{-\left(-2\right)±\sqrt{4+152}}{2\times 2}$$

Add 4 to 152.

$$x=\frac{-\left(-2\right)±\sqrt{156}}{2\times 2}$$

Take the square root of 156.

$$x=\frac{-\left(-2\right)±2\sqrt{39}}{2\times 2}$$

The opposite of -2 is 2.

$$x=\frac{2±2\sqrt{39}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{2±2\sqrt{39}}{4}$$

Now solve the equation x=\frac{2±2\sqrt{39}}{4} when ± is plus. Add 2 to 2\sqrt{39}.

$$x=\frac{2\sqrt{39}+2}{4}$$

Divide 2+2\sqrt{39} by 4.

$$x=\frac{\sqrt{39}+1}{2}$$

Now solve the equation x=\frac{2±2\sqrt{39}}{4} when ± is minus. Subtract 2\sqrt{39} from 2.

$$x=\frac{2-2\sqrt{39}}{4}$$

Divide 2-2\sqrt{39} by 4.

$$x=\frac{1-\sqrt{39}}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1+\sqrt{39}}{2} for x_{1} and \frac{1-\sqrt{39}}{2} for x_{2}.

$$2x^{2}-2x-19=2\left(x-\frac{\sqrt{39}+1}{2}\right)\left(x-\frac{1-\sqrt{39}}{2}\right)$$

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

$$x ^ 2 -1x -\frac{19}{2} = 0$$

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

$$r + s = 1 rs = -\frac{19}{2}$$

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

$$r = \frac{1}{2} - u s = \frac{1}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{19}{2}

$$(\frac{1}{2} - u) (\frac{1}{2} + u) = -\frac{19}{2}$$

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

$$\frac{1}{4} - u^2 = -\frac{19}{2}$$

Simplify the expression by subtracting \frac{1}{4} on both sides

$$-u^2 = -\frac{19}{2}-\frac{1}{4} = -\frac{39}{4}$$

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

$$u^2 = \frac{39}{4} u = \pm\sqrt{\frac{39}{4}} = \pm \frac{\sqrt{39}}{2}$$

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

$$r =\frac{1}{2} - \frac{\sqrt{39}}{2} = -2.622 s = \frac{1}{2} + \frac{\sqrt{39}}{2} = 3.622$$

Evaluate

$$2x^{2}-2x-19$$