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

Solution for (6x-4)(3x+5)=0 equation:

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

We multiply parentheses ..

(+18x^2+30x-12x-20)=0

We get rid of parentheses

18x^2+30x-12x-20=0

We add all the numbers together, and all the variables

18x^2+18x-20=0

a = 18; b = 18; c = -20;
Δ = b2-4ac
Δ = 182-4·18·(-20)
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-42}{2*18}=\frac{-60}{36}
=-1+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+42}{2*18}=\frac{24}{36}
=2/3
$