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

Solution for 3x^2+4x-1=x^2-3x-4 equation:

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2x^2+7x+3=0

a = 2; b = 7; c = +3;
Δ = b2-4ac
Δ = 72-4·2·3
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{-(7)-5}{2*2}=\frac{-12}{4}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+5}{2*2}=\frac{-2}{4}
=-1/2
$