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

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

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

We multiply parentheses

5x^2-2x^2-20x+6x+15=0

We add all the numbers together, and all the variables

3x^2-14x+15=0

a = 3; b = -14; c = +15;
Δ = b2-4ac
Δ = -142-4·3·15
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-4}{2*3}=\frac{10}{6}
=1+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+4}{2*3}=\frac{18}{6}
=3
$