9x^2-5x=x^2-4x

Solution for 9x^2-5x=x^2-4x equation:

9x^2-5x=x^2-4x

We move all terms to the left:

9x^2-5x-(x^2-4x)=0

We get rid of parentheses

9x^2-x^2-5x+4x=0

We add all the numbers together, and all the variables

8x^2-1x=0

a = 8; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·8·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*8}=\frac{0}{16}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*8}=\frac{2}{16}
=1/8
$