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

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

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

We multiply parentheses

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

We add all the numbers together, and all the variables

4x^2+3x-1=0

a = 4; b = 3; c = -1;
Δ = b2-4ac
Δ = 32-4·4·(-1)
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{-(3)-5}{2*4}=\frac{-8}{8}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+5}{2*4}=\frac{2}{8}
=1/4
$