(3x^2-4x+6)-(-2x^2+4)+(-5x-3)=1

Solution for (3x^2-4x+6)-(-2x^2+4)+(-5x-3)=1 equation:

(3x^2-4x+6)-(-2x^2+4)+(-5x-3)=1

We move all terms to the left:

(3x^2-4x+6)-(-2x^2+4)+(-5x-3)-(1)=0

We get rid of parentheses

2x^2+3x^2-4x-5x-4+6-3-1=0

We add all the numbers together, and all the variables

5x^2-9x-2=0

a = 5; b = -9; c = -2;
Δ = b2-4ac
Δ = -92-4·5·(-2)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-11}{2*5}=\frac{-2}{10}
=-1/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+11}{2*5}=\frac{20}{10}
=2
$