F(x)=3x(kwadrat) +5x-2 oblicz delte,x1,x2,xw.

3x² +5x-2 a=3 b=5 c=(-2) Δ=b²-4ac Δ=25+24 Δ=49 √Δ=7 x₁= -b-√Δ/2a = -5-7/6 = -12/6 = -2 x₂= -b+Δ/2a= -5+7/6 = 2/6 = ⅓

3x²+5x-2 a=3 b=5 c=-2 Δ=-b²-4 *a*c Δ=(-5)²-4*3*(-2) Δ=25+24 Δ=49 √Δ=7 x1=-b-√Δ/2*a=-5-7/2*3=-12/6=-2 x2=-b+√Δ/2*a=-5+7/2*3=2/6=1/3