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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-10x^2+13x+3=0

a = -10; b = 13; c = +3;
Δ = b2-4ac
Δ = 132-4·(-10)·3
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-17}{2*-10}=\frac{-30}{-20}
=1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+17}{2*-10}=\frac{4}{-20}
=-1/5
$