x+(20x2)=10x

Solution for x+(20x2)=10x equation:

x+(20x^2)=10x

We move all terms to the left:

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

determiningTheFunctionDomain
20x^2+x-10x=0

We add all the numbers together, and all the variables

20x^2-9x=0

a = 20; b = -9; c = 0;
Δ = b2-4ac
Δ = -92-4·20·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-9}{2*20}=\frac{0}{40}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+9}{2*20}=\frac{18}{40}
=9/20
$