-x^2-6x+3x^2+3x+10=3x^2

Solution for -x^2-6x+3x^2+3x+10=3x^2 equation:

-x^2-6x+3x^2+3x+10=3x^2

We move all terms to the left:

-x^2-6x+3x^2+3x+10-(3x^2)=0

determiningTheFunctionDomain
-x^2+3x^2-3x^2-6x+3x+10=0

We add all the numbers together, and all the variables

-1x^2-3x+10=0

a = -1; b = -3; c = +10;
Δ = b2-4ac
Δ = -32-4·(-1)·10
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{-(-3)-7}{2*-1}=\frac{-4}{-2}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+7}{2*-1}=\frac{10}{-2}
=-5
$