2x^2+4x-39=x^2

Solution for 2x^2+4x-39=x^2 equation:

2x^2+4x-39=x^2

We move all terms to the left:

2x^2+4x-39-(x^2)=0

determiningTheFunctionDomain
2x^2-x^2+4x-39=0

We add all the numbers together, and all the variables

x^2+4x-39=0

a = 1; b = 4; c = -39;
Δ = b2-4ac
Δ = 42-4·1·(-39)
Δ = 172
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{172}=\sqrt{4*43}=\sqrt{4}*\sqrt{43}=2\sqrt{43}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{43}}{2*1}=\frac{-4-2\sqrt{43}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{43}}{2*1}=\frac{-4+2\sqrt{43}}{2}
$