208=(10+2x)(x)

Solution for 208=(10+2x)(x) equation:

208=(10+2x)(x)

We move all terms to the left:

208-((10+2x)(x))=0

We add all the numbers together, and all the variables

-((2x+10)x)+208=0


We calculate terms in parentheses: -((2x+10)x), so:

(2x+10)x

We multiply parentheses

2x^2+10x

Back to the equation:

-(2x^2+10x)


We get rid of parentheses

-2x^2-10x+208=0

a = -2; b = -10; c = +208;
Δ = b2-4ac
Δ = -102-4·(-2)·208
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-42}{2*-2}=\frac{-32}{-4}
=+8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+42}{2*-2}=\frac{52}{-4}
=-13
$