2x*(x+3)=2*(3x+5)-x

Solution for 2x*(x+3)=2*(3x+5)-x equation:

2x(x+3)=2(3x+5)-x

We move all terms to the left:

2x(x+3)-(2(3x+5)-x)=0

We multiply parentheses

2x^2+6x-(2(3x+5)-x)=0


We calculate terms in parentheses: -(2(3x+5)-x), so:

2(3x+5)-x

We add all the numbers together, and all the variables

-1x+2(3x+5)

We multiply parentheses

-1x+6x+10

We add all the numbers together, and all the variables

5x+10

Back to the equation:

-(5x+10)


We get rid of parentheses

2x^2+6x-5x-10=0

We add all the numbers together, and all the variables

2x^2+x-10=0

a = 2; b = 1; c = -10;
Δ = b2-4ac
Δ = 12-4·2·(-10)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-9}{2*2}=\frac{-10}{4}
=-2+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+9}{2*2}=\frac{8}{4}
=2
$