X2+(x+1)2+(x+2)2=149

Solution for X2+(x+1)2+(x+2)2=149 equation:

X2+(X+1)2+(X+2)2=149

We move all terms to the left:

X2+(X+1)2+(X+2)2-(149)=0

We add all the numbers together, and all the variables

X^2+(X+1)2+(X+2)2-149=0

We multiply parentheses

X^2+2X+2X+2+4-149=0

We add all the numbers together, and all the variables

X^2+4X-143=0

a = 1; b = 4; c = -143;
Δ = b2-4ac
Δ = 42-4·1·(-143)
Δ = 588
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:

$\sqrt{\Delta}=\sqrt{588}=\sqrt{196*3}=\sqrt{196}*\sqrt{3}=14\sqrt{3}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-14\sqrt{3}}{2*1}=\frac{-4-14\sqrt{3}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+14\sqrt{3}}{2*1}=\frac{-4+14\sqrt{3}}{2}
$