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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

3a^2-7a-20=0

a = 3; b = -7; c = -20;
Δ = b2-4ac
Δ = -72-4·3·(-20)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-17}{2*3}=\frac{-10}{6}
=-1+2/3
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+17}{2*3}=\frac{24}{6}
=4
$