(x2-3)+x+(x-3)=56

Solution for (x2-3)+x+(x-3)=56 equation:

(x2-3)+x+(x-3)=56

We move all terms to the left:

(x2-3)+x+(x-3)-(56)=0

We add all the numbers together, and all the variables

(+x^2-3)+x+(x-3)-56=0

We get rid of parentheses

x^2+x+x-3-3-56=0

We add all the numbers together, and all the variables

x^2+2x-62=0

a = 1; b = 2; c = -62;
Δ = b2-4ac
Δ = 22-4·1·(-62)
Δ = 252
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:

$\sqrt{\Delta}=\sqrt{252}=\sqrt{36*7}=\sqrt{36}*\sqrt{7}=6\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-6\sqrt{7}}{2*1}=\frac{-2-6\sqrt{7}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+6\sqrt{7}}{2*1}=\frac{-2+6\sqrt{7}}{2}
$