x(x+1)+(x+2)=63

Solution for x(x+1)+(x+2)=63 equation:

x(x+1)+(x+2)=63

We move all terms to the left:

x(x+1)+(x+2)-(63)=0

We multiply parentheses

x^2+x+(x+2)-63=0

We get rid of parentheses

x^2+x+x+2-63=0

We add all the numbers together, and all the variables

x^2+2x-61=0

a = 1; b = 2; c = -61;
Δ = b2-4ac
Δ = 22-4·1·(-61)
Δ = 248
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:

$\sqrt{\Delta}=\sqrt{248}=\sqrt{4*62}=\sqrt{4}*\sqrt{62}=2\sqrt{62}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{62}}{2*1}=\frac{-2-2\sqrt{62}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{62}}{2*1}=\frac{-2+2\sqrt{62}}{2}
$