3x(x+2)=(x+1)(x-5)

Solution for 3x(x+2)=(x+1)(x-5) equation:

3x(x+2)=(x+1)(x-5)

We move all terms to the left:

3x(x+2)-((x+1)(x-5))=0

We multiply parentheses

3x^2+6x-((x+1)(x-5))=0

We multiply parentheses ..

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


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

(+x^2-5x+x-5)

We get rid of parentheses

x^2-5x+x-5

We add all the numbers together, and all the variables

x^2-4x-5

Back to the equation:

-(x^2-4x-5)


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2x^2+10x+5=0

a = 2; b = 10; c = +5;
Δ = b2-4ac
Δ = 102-4·2·5
Δ = 60
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:

$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{15}}{2*2}=\frac{-10-2\sqrt{15}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{15}}{2*2}=\frac{-10+2\sqrt{15}}{4}
$