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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

5n^2+17n-12=0

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

$\sqrt{\Delta}=\sqrt{529}=23$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-23}{2*5}=\frac{-40}{10}
=-4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+23}{2*5}=\frac{6}{10}
=3/5
$