X^2+(x+1)(x+1)+(x+2)(x+2)=302

Solution for X^2+(x+1)(x+1)+(x+2)(x+2)=302 equation:

X^2+(X+1)(X+1)+(X+2)(X+2)=302

We move all terms to the left:

X^2+(X+1)(X+1)+(X+2)(X+2)-(302)=0

We multiply parentheses ..

X^2+(+X^2+X+X+1)+(X+2)(X+2)-302=0

We get rid of parentheses

X^2+X^2+X+X+(X+2)(X+2)+1-302=0

We multiply parentheses ..

X^2+X^2+(+X^2+2X+2X+4)+X+X+1-302=0

We add all the numbers together, and all the variables

2X^2+(+X^2+2X+2X+4)+2X-301=0

We get rid of parentheses

2X^2+X^2+2X+2X+2X+4-301=0

We add all the numbers together, and all the variables

3X^2+6X-297=0

a = 3; b = 6; c = -297;
Δ = b2-4ac
Δ = 62-4·3·(-297)
Δ = 3600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3600}=60$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-60}{2*3}=\frac{-66}{6}
=-11
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+60}{2*3}=\frac{54}{6}
=9
$