-y2+3y+5=0 решите пожалуйста

[latex]-y^2+3y+5=0 \\ D=3^2-4*5*(-1)=9+20=29 \\ \sqrt{D} = \sqrt{29} \\ \\ y_{1} = \frac{-3+ \sqrt{29} }{2*(-1)} = \frac{-3+ \sqrt{29} }{-2} = \frac{3- \sqrt{29} }{2} \\ \\ y_2= \frac{-3- \sqrt{29} }{2*(-1)} = \frac{-3- \sqrt{29} }{-2} =\frac{3+ \sqrt{29} }{2} \\ \\ OTBET:\frac{3+ \sqrt{29} }{2} ; \ \frac{3- \sqrt{29} }{2} [/latex]