(x+1)(4x-1)+(x+1)(5x+3)=0

Solution for (x+1)(4x-1)+(x+1)(5x+3)=0 equation:

(x+1)(4x-1)+(x+1)(5x+3)=0

We multiply parentheses ..

(+4x^2-1x+4x-1)+(x+1)(5x+3)=0

We get rid of parentheses

4x^2-1x+4x+(x+1)(5x+3)-1=0

We multiply parentheses ..

4x^2+(+5x^2+3x+5x+3)-1x+4x-1=0

We add all the numbers together, and all the variables

4x^2+(+5x^2+3x+5x+3)+3x-1=0

We get rid of parentheses

4x^2+5x^2+3x+5x+3x+3-1=0

We add all the numbers together, and all the variables

9x^2+11x+2=0

a = 9; b = 11; c = +2;
Δ = b2-4ac
Δ = 112-4·9·2
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-7}{2*9}=\frac{-18}{18}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+7}{2*9}=\frac{-4}{18}
=-2/9
$