(2x+3)(3x+4)=180

Solution for (2x+3)(3x+4)=180 equation:

(2x+3)(3x+4)=180

We move all terms to the left:

(2x+3)(3x+4)-(180)=0

We multiply parentheses ..

(+6x^2+8x+9x+12)-180=0

We get rid of parentheses

6x^2+8x+9x+12-180=0

We add all the numbers together, and all the variables

6x^2+17x-168=0

a = 6; b = 17; c = -168;
Δ = b2-4ac
Δ = 172-4·6·(-168)
Δ = 4321
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{-(17)-\sqrt{4321}}{2*6}=\frac{-17-\sqrt{4321}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+\sqrt{4321}}{2*6}=\frac{-17+\sqrt{4321}}{12}
$