(3x+2)(2x+3)=49x+6

Solution for (3x+2)(2x+3)=49x+6 equation:

(3x+2)(2x+3)=49x+6

We move all terms to the left:

(3x+2)(2x+3)-(49x+6)=0

We get rid of parentheses

(3x+2)(2x+3)-49x-6=0

We multiply parentheses ..

(+6x^2+9x+4x+6)-49x-6=0

We get rid of parentheses

6x^2+9x+4x-49x+6-6=0

We add all the numbers together, and all the variables

6x^2-36x=0

a = 6; b = -36; c = 0;
Δ = b2-4ac
Δ = -362-4·6·0
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-36}{2*6}=\frac{0}{12}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+36}{2*6}=\frac{72}{12}
=6
$