(x-20)+21+42+29+(x+14)+x(x-10)=1260

Solution for (x-20)+21+42+29+(x+14)+x(x-10)=1260 equation:

(x-20)+21+42+29+(x+14)+x(x-10)=1260

We move all terms to the left:

(x-20)+21+42+29+(x+14)+x(x-10)-(1260)=0

We add all the numbers together, and all the variables

(x-20)+(x+14)+x(x-10)-1168=0

We multiply parentheses

x^2+(x-20)+(x+14)-10x-1168=0

We get rid of parentheses

x^2+x+x-10x-20+14-1168=0

We add all the numbers together, and all the variables

x^2-8x-1174=0

a = 1; b = -8; c = -1174;
Δ = b2-4ac
Δ = -82-4·1·(-1174)
Δ = 4760
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:

$\sqrt{\Delta}=\sqrt{4760}=\sqrt{4*1190}=\sqrt{4}*\sqrt{1190}=2\sqrt{1190}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{1190}}{2*1}=\frac{8-2\sqrt{1190}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{1190}}{2*1}=\frac{8+2\sqrt{1190}}{2}
$