x^2+4-4x=6x+31

Solution for x^2+4-4x=6x+31 equation:

x^2+4-4x=6x+31

We move all terms to the left:

x^2+4-4x-(6x+31)=0

We get rid of parentheses

x^2-4x-6x-31+4=0

We add all the numbers together, and all the variables

x^2-10x-27=0

a = 1; b = -10; c = -27;
Δ = b2-4ac
Δ = -102-4·1·(-27)
Δ = 208
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:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-4\sqrt{13}}{2*1}=\frac{10-4\sqrt{13}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+4\sqrt{13}}{2*1}=\frac{10+4\sqrt{13}}{2}
$