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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2x^2+x-10=0

a = 2; b = 1; c = -10;
Δ = b2-4ac
Δ = 12-4·2·(-10)
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{-(1)-9}{2*2}=\frac{-10}{4}
=-2+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+9}{2*2}=\frac{8}{4}
=2
$