Solve 2a^2+6a-2

$$2 a ^ { 2 } + 6 a - 2$$

Factor

$$2\left(a-\frac{-\sqrt{13}-3}{2}\right)\left(a-\frac{\sqrt{13}-3}{2}\right)$$

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.

$$2a^{2}+6a-2=0$$

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.

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

Square 6.

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

Multiply -4 times 2.

$$a=\frac{-6±\sqrt{36-8\left(-2\right)}}{2\times 2}$$

Multiply -8 times -2.

$$a=\frac{-6±\sqrt{36+16}}{2\times 2}$$

Add 36 to 16.

$$a=\frac{-6±\sqrt{52}}{2\times 2}$$

Take the square root of 52.

$$a=\frac{-6±2\sqrt{13}}{2\times 2}$$

Multiply 2 times 2.

$$a=\frac{-6±2\sqrt{13}}{4}$$

Now solve the equation a=\frac{-6±2\sqrt{13}}{4} when ± is plus. Add -6 to 2\sqrt{13}.

$$a=\frac{2\sqrt{13}-6}{4}$$

Divide -6+2\sqrt{13} by 4.

$$a=\frac{\sqrt{13}-3}{2}$$

Now solve the equation a=\frac{-6±2\sqrt{13}}{4} when ± is minus. Subtract 2\sqrt{13} from -6.

$$a=\frac{-2\sqrt{13}-6}{4}$$

Divide -6-2\sqrt{13} by 4.

$$a=\frac{-\sqrt{13}-3}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-3+\sqrt{13}}{2} for x_{1} and \frac{-3-\sqrt{13}}{2} for x_{2}.

$$2a^{2}+6a-2=2\left(a-\frac{\sqrt{13}-3}{2}\right)\left(a-\frac{-\sqrt{13}-3}{2}\right)$$

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

$$x ^ 2 +3x -1 = 0$$

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

$$r + s = -3 rs = -1$$

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

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

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

$$(-\frac{3}{2} - u) (-\frac{3}{2} + u) = -1$$

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

$$\frac{9}{4} - u^2 = -1$$

Simplify the expression by subtracting \frac{9}{4} on both sides

$$-u^2 = -1-\frac{9}{4} = -\frac{13}{4}$$

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

$$u^2 = \frac{13}{4} u = \pm\sqrt{\frac{13}{4}} = \pm \frac{\sqrt{13}}{2}$$

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

$$r =-\frac{3}{2} - \frac{\sqrt{13}}{2} = -3.303 s = -\frac{3}{2} + \frac{\sqrt{13}}{2} = 0.303$$

Evaluate

$$2\left(a^{2}+3a-1\right)$$