6x+5=1+2(3x+2)6x+5=1+23x+2

Solution for 6x+5=1+2(3x+2)6x+5=1+23x+2 equation:

6x+5=1+2(3x+2)6x+5=1+23x+2

We move all terms to the left:

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


We calculate terms in parentheses: -(1+2(3x+2)6x+5), so:

1+2(3x+2)6x+5

determiningTheFunctionDomain
2(3x+2)6x+1+5

We add all the numbers together, and all the variables

2(3x+2)6x+6

We multiply parentheses

36x^2+24x+6

Back to the equation:

-(36x^2+24x+6)


We get rid of parentheses

-36x^2+6x-24x-6+5=0

We add all the numbers together, and all the variables

-36x^2-18x-1=0

a = -36; b = -18; c = -1;
Δ = b2-4ac
Δ = -182-4·(-36)·(-1)
Δ = 180
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:

$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-6\sqrt{5}}{2*-36}=\frac{18-6\sqrt{5}}{-72}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+6\sqrt{5}}{2*-36}=\frac{18+6\sqrt{5}}{-72}
$