3(x^2-3x+3)-2(2x^2+3x-5)=0

Solution for 3(x^2-3x+3)-2(2x^2+3x-5)=0 equation:

3(x^2-3x+3)-2(2x^2+3x-5)=0

We multiply parentheses

3x^2-4x^2-9x-6x+9+10=0

We add all the numbers together, and all the variables

-1x^2-15x+19=0

a = -1; b = -15; c = +19;
Δ = b2-4ac
Δ = -152-4·(-1)·19
Δ = 301
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-\sqrt{301}}{2*-1}=\frac{15-\sqrt{301}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{301}}{2*-1}=\frac{15+\sqrt{301}}{-2}
$