(3a-5)(1+4a)=0

Solution for (3a-5)(1+4a)=0 equation:

(3a-5)(1+4a)=0

We add all the numbers together, and all the variables

(3a-5)(4a+1)=0

We multiply parentheses ..

(+12a^2+3a-20a-5)=0

We get rid of parentheses

12a^2+3a-20a-5=0

We add all the numbers together, and all the variables

12a^2-17a-5=0

a = 12; b = -17; c = -5;
Δ = b2-4ac
Δ = -172-4·12·(-5)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-23}{2*12}=\frac{-6}{24}
=-1/4
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+23}{2*12}=\frac{40}{24}
=1+2/3
$