(5x+3)*(x-4)=0

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

5x^2-17x-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:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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