Rozwiąż równanie: 3(x+2)²+(2x-1)²-7(x+3)(x-3) = 28 x-3/2+x = 4x-1/2-0,25(2x+3) x-3/2+x=4x-1/2-0,25(2x+3)

3(x+2)²+(2x-1)²-7(x+3)(x-3) = 28 3(x²+4x+4)+(4x²-4x+1)-7(x²-3²)=28 3x²+12x+12+4x²-4x+1-7x²+63=28 8x=-48 /:8 x=-6 x-3/2+x = 4x-1/2-0,25(2x+3) (x-3)*[2-0,25(2x+3)]=(2+x)(4x-1) (x-3)(2-0,5x-0,75)=8x-2+4x²-x (x-3)(1,25-0,5x)=4x²+7x-2 1,25x-0,5x²-3,75+1,5x=4x²+7x-2 -4,5x²-4,25x-1,75 =0 i dalej trzeba liczyc delte, x₁ i x₂

3(x+2)²+(2x-1)²-7(x+3)(x-3) = 28 3(x²+2x+4)+4x²-x+1-7(x²-3x+3x-9)=28 3x²+6x+12+4x²-x+1-7x²+21x-21x+63=28 5x=-48 x=-9,6 x-3/2+x = 4x-1/2-0,25(2x+3) 2x-3/2=4x-1/2-0.5-0.75 2x-4x+0,5x=-1/2-1,25+3/2 -1,5x=1/4 x=-0,16 ;)