3(x2+4)+5x=2x2+6(x+3)

Solution for 3(x2+4)+5x=2x2+6(x+3) equation:

3(x2+4)+5x=2x^2+6(x+3)

We move all terms to the left:

3(x2+4)+5x-(2x^2+6(x+3))=0

We add all the numbers together, and all the variables

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

We multiply parentheses

3x^2+5x-(2x^2+6(x+3))+12=0


We calculate terms in parentheses: -(2x^2+6(x+3)), so:

2x^2+6(x+3)

We multiply parentheses

2x^2+6x+18

Back to the equation:

-(2x^2+6x+18)


We get rid of parentheses

3x^2-2x^2+5x-6x-18+12=0

We add all the numbers together, and all the variables

x^2-1x-6=0

a = 1; b = -1; c = -6;
Δ = b2-4ac
Δ = -12-4·1·(-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{-(-1)-5}{2*1}=\frac{-4}{2}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+5}{2*1}=\frac{6}{2}
=3
$