Algebra

\(2x^2+4x-1=x^2-8x+3\)

We can make this is the form  

\(ax^2+bx+c=0\)

We get:

\(x^2+12x-4=0\)

a = 1, b = 12, and c = -4.

Using the quadratic formula \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\), we get:

\(x = {-12 \pm \sqrt{12^2+16} \over 2}\)

\(x = {-12 \pm \sqrt{160} \over 2}\)

\(x = {-12 - \sqrt{160} \over 2}\), \(x = {-12 + \sqrt{160} \over 2}\)

Now we can square both values of x and then add them:

-151.894663844 + 0.10533615595 = -151.789327688

Answer: -151.789327688

Re-arrange to 

x^2 +12x -4 = 0     complete the square 

(x+6)^2  = 4 + 36 

x+6 = +- sqrt 40 

x =  - 6 +- sqrt 40      <==== these are the roots 

Now square the roots

(-6 + sqrt 40)² = 76 - 12 sqrt 40 

(-6  - sqrt 40)²  = 76 +12 sqrt 40

                        added together =  76 + 76 =   + 152  

Simplify as

x^2 + 12x - 4 =  0

Call the roots a,b

Product of  roots

ab  = -4

2ab  =  -8

Sum of  roots 

a + b  =  -12      square both sides

a^2 + 2ab + b^2 = 144

a^2 + b^2  -8   =144

a^2 + b^2   = 152

PLug in the quadratic formula:\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)