10^2=x(2x+4)

Solution for 10^2=x(2x+4) equation:

10^2=x(2x+4)

We move all terms to the left:

10^2-(x(2x+4))=0

We add all the numbers together, and all the variables

-(x(2x+4))+100=0


We calculate terms in parentheses: -(x(2x+4)), so:

x(2x+4)

We multiply parentheses

2x^2+4x

Back to the equation:

-(2x^2+4x)


We get rid of parentheses

-2x^2-4x+100=0

a = -2; b = -4; c = +100;
Δ = b2-4ac
Δ = -42-4·(-2)·100
Δ = 816
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:

$\sqrt{\Delta}=\sqrt{816}=\sqrt{16*51}=\sqrt{16}*\sqrt{51}=4\sqrt{51}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{51}}{2*-2}=\frac{4-4\sqrt{51}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{51}}{2*-2}=\frac{4+4\sqrt{51}}{-4}
$