x²= ( 7-x) a drugi to x²-2+x=4x-x²-3 Proszę o pomoc .... dzięki z góry :)

1) x²=7-x x²+x-7=0 Δ=1²-4*1*(-7)=1+28=29 √Δ=√29 x1=(-1+√29)/2 x2=(-1-√29)/2 2) x²-2+x=4x-x²-3 x²-2+x-4x+x²+3=0 2x²-3x+1=0 Δ=(-3)²-4*1*2=9-8=1 √Δ=1 x1=(3+1)/2*2=4/4=1 x2=(3-1)/2*2=2/4=1/2

[latex] x^{2} =(7-x) [/latex] [latex]x^{2} -(7-x)=0[/latex] [latex] x^{2}+x-7=0[/latex] Δ=1+28=29 √Δ=√29 [latex]x1= frac{1- sqrt{29} }{2} [/latex] [latex]x2= frac{1+ sqrt{29} }{2} [/latex] [latex] x^{2} -2+x=4x- x^{2} -3[/latex] [latex]2 x^{2} -3x+1=0[/latex] Δ=9-8=1 √Δ=1 [latex]x1= frac{3-1}{4}= frac{1}{2} [/latex] [latex]x2= frac{3+1}{4}=1[/latex]