(4x-3)(5x+2)=3

Solution for (4x-3)(5x+2)=3 equation:

(4x-3)(5x+2)=3

We move all terms to the left:

(4x-3)(5x+2)-(3)=0

We multiply parentheses ..

(+20x^2+8x-15x-6)-3=0

We get rid of parentheses

20x^2+8x-15x-6-3=0

We add all the numbers together, and all the variables

20x^2-7x-9=0

a = 20; b = -7; c = -9;
Δ = b2-4ac
Δ = -72-4·20·(-9)
Δ = 769
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{769}}{2*20}=\frac{7-\sqrt{769}}{40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{769}}{2*20}=\frac{7+\sqrt{769}}{40}
$