20^2=x(42+x)

Solution for 20^2=x(42+x) equation:

20^2=x(42+x)

We move all terms to the left:

20^2-(x(42+x))=0

We add all the numbers together, and all the variables

-(x(x+42))+20^2=0

We add all the numbers together, and all the variables

-(x(x+42))+400=0


We calculate terms in parentheses: -(x(x+42)), so:

x(x+42)

We multiply parentheses

x^2+42x

Back to the equation:

-(x^2+42x)


We get rid of parentheses

-x^2-42x+400=0

We add all the numbers together, and all the variables

-1x^2-42x+400=0

a = -1; b = -42; c = +400;
Δ = b2-4ac
Δ = -422-4·(-1)·400
Δ = 3364
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3364}=58$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-58}{2*-1}=\frac{-16}{-2}
=+8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+58}{2*-1}=\frac{100}{-2}
=-50
$