(5x+1)(3x-2)=0

Solution for (5x+1)(3x-2)=0 equation:

(5x+1)(3x-2)=0

We multiply parentheses ..

(+15x^2-10x+3x-2)=0

We get rid of parentheses

15x^2-10x+3x-2=0

We add all the numbers together, and all the variables

15x^2-7x-2=0

a = 15; b = -7; c = -2;
Δ = b2-4ac
Δ = -72-4·15·(-2)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-13}{2*15}=\frac{-6}{30}
=-1/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+13}{2*15}=\frac{20}{30}
=2/3
$