437=(21+x)(21-x)

Solution for 437=(21+x)(21-x) equation:

437=(21+x)(21-x)

We move all terms to the left:

437-((21+x)(21-x))=0

We add all the numbers together, and all the variables

-((x+21)(-1x+21))+437=0

We multiply parentheses ..

-((-1x^2+21x-21x+441))+437=0


We calculate terms in parentheses: -((-1x^2+21x-21x+441)), so:

(-1x^2+21x-21x+441)

We get rid of parentheses

-1x^2+21x-21x+441

We add all the numbers together, and all the variables

-1x^2+441

Back to the equation:

-(-1x^2+441)


We get rid of parentheses

1x^2-441+437=0

We add all the numbers together, and all the variables

x^2-4=0

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

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