221^2=(x+4)x

Solution for 221^2=(x+4)x equation:

221^2=(x+4)x

We move all terms to the left:

221^2-((x+4)x)=0

We add all the numbers together, and all the variables

-((x+4)x)+48841=0


We calculate terms in parentheses: -((x+4)x), so:

(x+4)x

We multiply parentheses

x^2+4x

Back to the equation:

-(x^2+4x)


We get rid of parentheses

-x^2-4x+48841=0

We add all the numbers together, and all the variables

-1x^2-4x+48841=0

a = -1; b = -4; c = +48841;
Δ = b2-4ac
Δ = -42-4·(-1)·48841
Δ = 195380
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:

$\sqrt{\Delta}=\sqrt{195380}=\sqrt{4*48845}=\sqrt{4}*\sqrt{48845}=2\sqrt{48845}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{48845}}{2*-1}=\frac{4-2\sqrt{48845}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{48845}}{2*-1}=\frac{4+2\sqrt{48845}}{-2}
$