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 - y^2 = 13
y^2- 6y + 9 + 10y - y^2 = 13
4y = 13 - 9
4y = 4                                /:4
y = 1  

x^2 + x = (x - 3)(x + 3)
x^2 + x = x^2 - 9
x^2- x^2 + x = -9
x = -9    

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