2x^2+4x-(x^2-3x)+6=0

Solution for 2x^2+4x-(x^2-3x)+6=0 equation:

2x^2+4x-(x^2-3x)+6=0

We get rid of parentheses

2x^2-x^2+4x+3x+6=0

We add all the numbers together, and all the variables

x^2+7x+6=0

a = 1; b = 7; c = +6;
Δ = b2-4ac
Δ = 72-4·1·6
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*1}=\frac{-12}{2}
=-6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+5}{2*1}=\frac{-2}{2}
=-1
$