10x^2=41x

Solution for 10x^2=41x equation:

10x^2=41x

We move all terms to the left:

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