Question

(y-3)²+10y-y²=13
x²+x=(x-3)*(x+3)
(3x+1)=(2x-1)²+5x²

Answer 1

(y - 3)² + 10y - y² = 13
y² - 6y + 9 + 10y - y² = 13
4y = 13 - 9
4y = 4    /:4
y = 1 


x² + x = (x - 3)(x + 3)
x² + x = x² - 9
x² - x² + x = -9
x = -9   


(3x + 1)² = (2x - 1)² + 5x²
9x² + 6x + 1 = 4x² - 4x + 1 + 5x²
9x² + 6x + 1 = 9x² - 4x + 1
9x² - 9x² + 6x + 4x = 1 - 1
10x = 0
x = 0