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

Solution for x^2-4x=-6x^2+3x equation:

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

7x^2-7x=0

a = 7; b = -7; c = 0;
Δ = b2-4ac
Δ = -72-4·7·0
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{-(-7)-7}{2*7}=\frac{0}{14}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+7}{2*7}=\frac{14}{14}
=1
$