(x-7)2+x^2=(x+1)2

Solution for (x-7)2+x^2=(x+1)2 equation:

(x-7)2+x^2=(x+1)2

We move all terms to the left:

(x-7)2+x^2-((x+1)2)=0

We multiply parentheses

x^2+2x-((x+1)2)-14=0


We calculate terms in parentheses: -((x+1)2), so:

(x+1)2

We multiply parentheses

2x+2

Back to the equation:

-(2x+2)


We get rid of parentheses

x^2+2x-2x-2-14=0

We add all the numbers together, and all the variables

x^2-16=0

a = 1; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·1·(-16)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8}{2*1}=\frac{-8}{2}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8}{2*1}=\frac{8}{2}
=4
$