x2+(x2+1)=365

Solution for x2+(x2+1)=365 equation:

x2+(x2+1)=365

We move all terms to the left:

x2+(x2+1)-(365)=0

We add all the numbers together, and all the variables

(+x^2+1)+x2-365=0

We add all the numbers together, and all the variables

x^2+(+x^2+1)-365=0

We get rid of parentheses

x^2+x^2+1-365=0

We add all the numbers together, and all the variables

2x^2-364=0

a = 2; b = 0; c = -364;
Δ = b2-4ac
Δ = 02-4·2·(-364)
Δ = 2912
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:

$\sqrt{\Delta}=\sqrt{2912}=\sqrt{16*182}=\sqrt{16}*\sqrt{182}=4\sqrt{182}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{182}}{2*2}=\frac{0-4\sqrt{182}}{4}
=-\frac{4\sqrt{182}}{4}
=-\sqrt{182}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{182}}{2*2}=\frac{0+4\sqrt{182}}{4}
=\frac{4\sqrt{182}}{4}
=\sqrt{182}
$