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

Solution for (x-3)(2x+1)+(2x+1)(4x+3)=0 equation:

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

We multiply parentheses ..

(+2x^2+x-6x-3)+(2x+1)(4x+3)=0

We get rid of parentheses

2x^2+x-6x+(2x+1)(4x+3)-3=0

We multiply parentheses ..

2x^2+(+8x^2+6x+4x+3)+x-6x-3=0

We add all the numbers together, and all the variables

2x^2+(+8x^2+6x+4x+3)-5x-3=0

We get rid of parentheses

2x^2+8x^2+6x+4x-5x+3-3=0

We add all the numbers together, and all the variables

10x^2+5x=0

a = 10; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·10·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*10}=\frac{-10}{20}
=-1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*10}=\frac{0}{20}
=0
$