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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

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


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

3(x^2+2)

We multiply parentheses

3x^2+6

Back to the equation:

-(3x^2+6)


We add all the numbers together, and all the variables

2x^2+7x-(3x^2+6)=0

We get rid of parentheses

2x^2-3x^2+7x-6=0

We add all the numbers together, and all the variables

-1x^2+7x-6=0

a = -1; b = 7; c = -6;
Δ = b2-4ac
Δ = 72-4·(-1)·(-6)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-5}{2*-1}=\frac{-12}{-2}
=+6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+5}{2*-1}=\frac{-2}{-2}
=1
$