41x=10x^2

Solution for 41x=10x^2 equation:

41x=10x^2

We move all terms to the left:

41x-(10x^2)=0

determiningTheFunctionDomain
-10x^2+41x=0

a = -10; b = 41; c = 0;
Δ = b2-4ac
Δ = 412-4·(-10)·0
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-41}{2*-10}=\frac{-82}{-20}
=4+1/10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+41}{2*-10}=\frac{0}{-20}
=0
$