(3x^2)=2x(10+2x)2

Solution for (3x^2)=2x(10+2x)2 equation:

(3x^2)=2x(10+2x)2

We move all terms to the left:

(3x^2)-(2x(10+2x)2)=0

We add all the numbers together, and all the variables

3x^2-(2x(2x+10)2)=0


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

2x(2x+10)2

We multiply parentheses

8x^2+40x

Back to the equation:

-(8x^2+40x)


We get rid of parentheses

3x^2-8x^2-40x=0

We add all the numbers together, and all the variables

-5x^2-40x=0

a = -5; b = -40; c = 0;
Δ = b2-4ac
Δ = -402-4·(-5)·0
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-40}{2*-5}=\frac{0}{-10}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+40}{2*-5}=\frac{80}{-10}
=-8
$