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

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

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2+7x=0

a = 1; b = 7; c = 0;
Δ = b2-4ac
Δ = 72-4·1·0
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{-(7)-7}{2*1}=\frac{-14}{2}
=-7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+7}{2*1}=\frac{0}{2}
=0
$