4x+3(3x+1)=6x(2x+1)

Solution for 4x+3(3x+1)=6x(2x+1) equation:

4x+3(3x+1)=6x(2x+1)

We move all terms to the left:

4x+3(3x+1)-(6x(2x+1))=0

We multiply parentheses

4x+9x-(6x(2x+1))+3=0


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

6x(2x+1)

We multiply parentheses

12x^2+6x

Back to the equation:

-(12x^2+6x)


We add all the numbers together, and all the variables

13x-(12x^2+6x)+3=0

We get rid of parentheses

-12x^2+13x-6x+3=0

We add all the numbers together, and all the variables

-12x^2+7x+3=0

a = -12; b = 7; c = +3;
Δ = b2-4ac
Δ = 72-4·(-12)·3
Δ = 193
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{193}}{2*-12}=\frac{-7-\sqrt{193}}{-24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{193}}{2*-12}=\frac{-7+\sqrt{193}}{-24}
$