10x^2-20x+10=9x

Solution for 10x^2-20x+10=9x equation:

10x^2-20x+10=9x

We move all terms to the left:

10x^2-20x+10-(9x)=0

We add all the numbers together, and all the variables

10x^2-29x+10=0

a = 10; b = -29; c = +10;
Δ = b2-4ac
Δ = -292-4·10·10
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-21}{2*10}=\frac{8}{20}
=2/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+21}{2*10}=\frac{50}{20}
=2+1/2
$