50.5=(2x^2)/2-3x+x^2

Solution for 50.5=(2x^2)/2-3x+x^2 equation:

50.5=(2x^2)/2-3x+x^2

We move all terms to the left:

50.5-((2x^2)/2-3x+x^2)=0


Domain of the equation: 2-3x+x^2)!=0

We move all terms containing x to the left, all other terms to the right

x^2)-3x!=-2

x∈R


We get rid of parentheses

-2x^2/2-x^2+3x+50.5=0

We multiply all the terms by the denominator


-2x^2-x^2*2+3x*2+(50.5)*2=0

We add all the numbers together, and all the variables

-2x^2-x^2*2+3x*2+101=0

Wy multiply elements

-2x^2-2x^2+6x+101=0

We add all the numbers together, and all the variables

-4x^2+6x+101=0

a = -4; b = 6; c = +101;
Δ = b2-4ac
Δ = 62-4·(-4)·101
Δ = 1652
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1652}=\sqrt{4*413}=\sqrt{4}*\sqrt{413}=2\sqrt{413}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{413}}{2*-4}=\frac{-6-2\sqrt{413}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{413}}{2*-4}=\frac{-6+2\sqrt{413}}{-8}
$