3x(2x+3)=2x(x+4,5)+2

Solution for 3x(2x+3)=2x(x+4,5)+2 equation:

3x(2x+3)=2x(x+4.5)+2

We move all terms to the left:

3x(2x+3)-(2x(x+4.5)+2)=0

We multiply parentheses

6x^2+9x-(2x(x+4.5)+2)=0


We calculate terms in parentheses: -(2x(x+4.5)+2), so:

2x(x+4.5)+2

We multiply parentheses

2x^2+9x+2

Back to the equation:

-(2x^2+9x+2)


We get rid of parentheses

6x^2-2x^2+9x-9x-2=0

We add all the numbers together, and all the variables

4x^2-2=0

a = 4; b = 0; c = -2;
Δ = b2-4ac
Δ = 02-4·4·(-2)
Δ = 32
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:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{2}}{2*4}=\frac{0-4\sqrt{2}}{8}
=-\frac{4\sqrt{2}}{8}
=-\frac{\sqrt{2}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{2}}{2*4}=\frac{0+4\sqrt{2}}{8}
=\frac{4\sqrt{2}}{8}
=\frac{\sqrt{2}}{2}
$