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

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

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

We multiply parentheses ..

(+16x^2+4x+4x+1)-3(4x+1)(x-3)=0

We get rid of parentheses

16x^2+4x+4x-3(4x+1)(x-3)+1=0

We multiply parentheses ..

16x^2-3(+4x^2-12x+x-3)+4x+4x+1=0

We add all the numbers together, and all the variables

16x^2-3(+4x^2-12x+x-3)+8x+1=0

We multiply parentheses

16x^2-12x^2+36x-3x+8x+9+1=0

We add all the numbers together, and all the variables

4x^2+41x+10=0

a = 4; b = 41; c = +10;
Δ = b2-4ac
Δ = 412-4·4·10
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-39}{2*4}=\frac{-80}{8}
=-10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+39}{2*4}=\frac{-2}{8}
=-1/4
$