How to solve the quadratic equation of one variable 3x * 2-5x + 2 = 0, ;

3x * 2-5x + 2 = 0 to get (3x-2) (x-1) = 0, so 3x-2 = 0 or X-1 = 0, so x = 2 / 3 or x = 1

（8X-4）^2+（5X-2）^2=(3X+2)^2
It's really bad luck to solve this equation today. I don't know where I got it wrong
（8X-4）^2+（5X-2）^2=(3X+2)^2

(8x-4)²=(3x+2)²-(5x-2)²
(8x-4)²=8x*(4-2x)
64x²-64x+16=32x-16x²
80x²-96x+16=0
5x²-6x+1=0
(5x-1)(x-1)=0
X = 1 / 5 or 1

Equations {5x + 8y + 12z = 100 x + y + Z = 14
Equations
｛5x+8y+12z=100
x+y+z=14

There will be infinitely many solutions
5x + 8y = 100 - 12z
x + y = 14 - z → 5x + 5y = 70 - 5z
3y = 30 - 7z
y = 10 - (7/3)z
x = 4 + (4/3)z = 4(1 + z/3)
Let x = 4N
z = 3n - 3
y = 10 - (7/3)(3n - 3) = 17 - 7n
So the solution is:
x = 4n
y = 17 - 7n
z = 3n - 3
N is any integer

How to solve 8y + 4.5X = L; 8y-4.5x = 18 (L + 18) / x = (L-18) y

Formula 1 + 2 gives L + 18 = 16y, formula 1-2 gives L-18 = 9x
Take the formula into the third formula, and you can find x = plus or minus 4, y = plus or minus 3
Then we find L

How to solve the quadratic equation of one variable 3x * 2-5x + 2 = 0, ;