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

Solution for (6x^2-2x+5)-(3x^2+3x+4)=0 equation:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

3x^2-5x+1=0

a = 3; b = -5; c = +1;
Δ = b2-4ac
Δ = -52-4·3·1
Δ = 13
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{-(-5)-\sqrt{13}}{2*3}=\frac{5-\sqrt{13}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{13}}{2*3}=\frac{5+\sqrt{13}}{6}
$