40-6x+x^2/3=40-12x+x^2

Solution for 40-6x+x^2/3=40-12x+x^2 equation:

40-6x+x^2/3=40-12x+x^2

We move all terms to the left:

40-6x+x^2/3-(40-12x+x^2)=0

We get rid of parentheses

x^2/3-x^2+12x-6x-40+40=0

We multiply all the terms by the denominator


x^2-x^2*3+12x*3-6x*3-40*3+40*3=0

We add all the numbers together, and all the variables

x^2-x^2*3+12x*3-6x*3=0

Wy multiply elements

x^2-3x^2+36x-18x=0

We add all the numbers together, and all the variables

-2x^2+18x=0

a = -2; b = 18; c = 0;
Δ = b2-4ac
Δ = 182-4·(-2)·0
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-18}{2*-2}=\frac{-36}{-4}
=+9
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+18}{2*-2}=\frac{0}{-4}
=0
$