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

Solution for (2x+1)(x-3)+(x+6)(2x+1)=0 equation:

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

We multiply parentheses ..

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

We get rid of parentheses

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

We multiply parentheses ..

2x^2+(+2x^2+x+12x+6)-6x+x-3=0

We add all the numbers together, and all the variables

2x^2+(+2x^2+x+12x+6)-5x-3=0

We get rid of parentheses

2x^2+2x^2+x+12x-5x+6-3=0

We add all the numbers together, and all the variables

4x^2+8x+3=0

a = 4; b = 8; c = +3;
Δ = b2-4ac
Δ = 82-4·4·3
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4}{2*4}=\frac{-12}{8}
=-1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4}{2*4}=\frac{-4}{8}
=-1/2
$