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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

4x^2+2x+6x-((3x+5))=0


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

(3x+5)

We get rid of parentheses

3x+5

Back to the equation:

-(3x+5)


We add all the numbers together, and all the variables

4x^2+8x-(3x+5)=0

We get rid of parentheses

4x^2+8x-3x-5=0

We add all the numbers together, and all the variables

4x^2+5x-5=0

a = 4; b = 5; c = -5;
Δ = b2-4ac
Δ = 52-4·4·(-5)
Δ = 105
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{105}}{2*4}=\frac{-5-\sqrt{105}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{105}}{2*4}=\frac{-5+\sqrt{105}}{8}
$