(3x+5)(4x-5)=40

Solution for (3x+5)(4x-5)=40 equation:

(3x+5)(4x-5)=40

We move all terms to the left:

(3x+5)(4x-5)-(40)=0

We multiply parentheses ..

(+12x^2-15x+20x-25)-40=0

We get rid of parentheses

12x^2-15x+20x-25-40=0

We add all the numbers together, and all the variables

12x^2+5x-65=0

a = 12; b = 5; c = -65;
Δ = b2-4ac
Δ = 52-4·12·(-65)
Δ = 3145
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{-(5)-\sqrt{3145}}{2*12}=\frac{-5-\sqrt{3145}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{3145}}{2*12}=\frac{-5+\sqrt{3145}}{24}
$