(4x-5)(3x)=6x+45

Solution for (4x-5)(3x)=6x+45 equation:

(4x-5)(3x)=6x+45

We move all terms to the left:

(4x-5)(3x)-(6x+45)=0

We multiply parentheses

12x^2-15x-(6x+45)=0

We get rid of parentheses

12x^2-15x-6x-45=0

We add all the numbers together, and all the variables

12x^2-21x-45=0

a = 12; b = -21; c = -45;
Δ = b2-4ac
Δ = -212-4·12·(-45)
Δ = 2601
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2601}=51$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-51}{2*12}=\frac{-30}{24}
=-1+1/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+51}{2*12}=\frac{72}{24}
=3
$