Question 389186
Quadratic First (~=square root)

x^(2)+8x-2=0

Use the quadratic formula to find the solutions.  In this case, the values are a=1, b=8, and c=-2.
x=(-b\~(b^(2)-4ac))/(2a) where ax^(2)+bx+c=0

Use the standard form of the equation to find a, b, and c for this quadratic.
a=1, b=8, and c=-2

Substitute in the values of a=1, b=8, and c=-2.
x=(-8\~((8)^(2)-4(1)(-2)))/(2(1))

Simplify the section inside the radical.
x=(-8\6~(2))/(2(1))

Simplify the denominator of the quadratic formula.
x=(-8\6~(2))/(2)

First, solve the + portion of +-.
x=(-8+6~(2))/(2)

Factor out the GCF of 2 from each term in the polynomial.
x=(2(-4)+2(3~(2)))/(2)

Factor out the GCF of 2 from -8+6~(2).
x=(2(-4+3~(2)))/(2)

Reduce the expression (2(-4+3~(2)))/(2) by removing a factor of 2 from the numerator and denominator.
x=(-4+3~(2))

Remove the parentheses around the expression -4+3~(2).
x=-4+3~(2)

Next, solve the - portion of +-.
x=(-8-6~(2))/(2)

Factor out the GCF of 2 from each term in the polynomial.
x=(2(-4)+2(-3~(2)))/(2)

Factor out the GCF of 2 from -8-6~(2).
x=(2(-4-3~(2)))/(2)

Reduce the expression (2(-4-3~(2)))/(2) by removing a factor of 2 from the numerator and denominator.
x=(-4-3~(2))

Remove the parentheses around the expression -4-3~(2).
x=-4-3~(2)

The final answer is the combination of both solutions.
x=-4+3~(2),-4-3~(2)_x=0.2426407,-8.24264


Now, completing the square


x^(2)+8x-2=0

Since -2 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 2 to both sides.
x^(2)+8x=2

To create a trinomial square on the left-hand side of the equation, add a value to both sides of the equation that is equal to the square of half the coefficient of x.  In this problem, add (4)^(2) to both sides of the equation.
x^(2)+8x+16=2+16

Add 16 to 2 to get 18.
x^(2)+8x+16=18

Factor the perfect trinomial square into (x+4)^(2).
(x+4)^(2)=18

Take the square root of each side of the equation to setup the solution for x.
~((x+4)^(2))=\~(18)

Remove the perfect root factor (x+4) under the radical to solve for x.
(x+4)=\~(18)

Pull all perfect square roots out from under the radical.  In this case, remove the 3 because it is a perfect square.
(x+4)=\3~(2)

First, substitute in the + portion of the \ to find the first solution.
(x+4)=3~(2)

Remove the parentheses around the expression x+4.
x+4=3~(2)

Since 4 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 4 from both sides.
x=-4+3~(2)

Move all terms not containing x to the right-hand side of the equation.
x=3~(2)-4

Next, substitute in the - portion of the \ to find the second solution.
(x+4)=-3~(2)

Remove the parentheses around the expression x+4.
x+4=-3~(2)

Since 4 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 4 from both sides.
x=-4+-3~(2)

Move all terms not containing x to the right-hand side of the equation.
x=-3~(2)-4

The complete solution is the result of both the + and - portions of the solution.
x=3~(2)-4,-3~(2)-4