2x(x+1)+4x=3(2x-1)+8

Solution for 2x(x+1)+4x=3(2x-1)+8 equation:

2x(x+1)+4x=3(2x-1)+8

We move all terms to the left:

2x(x+1)+4x-(3(2x-1)+8)=0

We add all the numbers together, and all the variables

4x+2x(x+1)-(3(2x-1)+8)=0

We multiply parentheses

2x^2+4x+2x-(3(2x-1)+8)=0


We calculate terms in parentheses: -(3(2x-1)+8), so:

3(2x-1)+8

We multiply parentheses

6x-3+8

We add all the numbers together, and all the variables

6x+5

Back to the equation:

-(6x+5)


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2x^2-5=0

a = 2; b = 0; c = -5;
Δ = b2-4ac
Δ = 02-4·2·(-5)
Δ = 40
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:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{10}}{2*2}=\frac{0-2\sqrt{10}}{4}
=-\frac{2\sqrt{10}}{4}
=-\frac{\sqrt{10}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{10}}{2*2}=\frac{0+2\sqrt{10}}{4}
=\frac{2\sqrt{10}}{4}
=\frac{\sqrt{10}}{2}
$