rozwiąż równanie : a) (x-3) (x+2) = (x+4) (2-x) - 14 b) (2-3x)do kwadratu + (3+x) (x-1) = 1+x c) (x-3)do kwadratu + (x+5)do kwadratu = 34-x

a) (x-3)(x+2) = (x+4)(2-x)-14 x²+2x-3x-6 = 2x-x²+8-4x-14 x²-x-6 = -x²-2x-6 x²+x²-x+2x-6+6 = 0 2x²+x = 0 Δ = 1²-4*2*0 Δ = 1-0 Δ = 1 √Δ = 1 x1 = (-1-1)/(2*2) x1 = -2/4 x1 = -0,5 x2 = (-1+1)/(2*2) x2 = 0/4 x2 = 0   b) (2-3x)²+(3+x)(x-1) = 1+x 4-12x+9x²+3x-3+x²-x = 1+x 10x²-10x+1 = 1+x 10x²-11x = 0 Δ = (-11)²-4*10*0 Δ = 121-0 Δ = 121 √Δ = 11 x1 = [-(-11)-11]/(2*10) x1 = (11-11)/20 x1 = 0/20 x1 = 0 x2 = [-(-11)+11]/(2*10) x2 = (11+11)/20 x2 = 22/20 x2 = 1,1   c) (x-3)²+(x+5)² = 34-x x²-6x+9+x²+10x+25 = 34-x 2x²+4x+34 = 34-x 2x²+5x = 0 Δ = 5²-4*2*0 Δ = 25-0 Δ = 25 √Δ = 5 x1 = (-5-5)/(2*2) x1 = -10/4 x1 = -2,5 x2 = (-5+5)/(2*2) x2 = 0/4 x2 = 0