(4x+1)(x-3)=0

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

(4x+1)(x-3)=0

We multiply parentheses ..

(+4x^2-12x+x-3)=0

We get rid of parentheses

4x^2-12x+x-3=0

We add all the numbers together, and all the variables

4x^2-11x-3=0

a = 4; b = -11; c = -3;
Δ = b2-4ac
Δ = -112-4·4·(-3)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-13}{2*4}=\frac{-2}{8}
=-1/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+13}{2*4}=\frac{24}{8}
=3
$