2x^2-4x=10x^2-5x

Solution for 2x^2-4x=10x^2-5x equation:

2x^2-4x=10x^2-5x

We move all terms to the left:

2x^2-4x-(10x^2-5x)=0

We get rid of parentheses

2x^2-10x^2-4x+5x=0

We add all the numbers together, and all the variables

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