(5x^2-3)+(2x^2-3x)=x^2

Solution for (5x^2-3)+(2x^2-3x)=x^2 equation:

(5x^2-3)+(2x^2-3x)=x^2

We move all terms to the left:

(5x^2-3)+(2x^2-3x)-(x^2)=0

determiningTheFunctionDomain
-x^2+(5x^2-3)+(2x^2-3x)=0

We add all the numbers together, and all the variables

-1x^2+(5x^2-3)+(2x^2-3x)=0

We get rid of parentheses

-1x^2+5x^2+2x^2-3x-3=0

We add all the numbers together, and all the variables

6x^2-3x-3=0

a = 6; b = -3; c = -3;
Δ = b2-4ac
Δ = -32-4·6·(-3)
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{-(-3)-9}{2*6}=\frac{-6}{12}
=-1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+9}{2*6}=\frac{12}{12}
=1
$