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$$\left.100x\cdot(x+3)=559\quad\right|\text{Klammer ausmultiplizieren}$$$$\left.100x^2+300x=559\quad\right|-559$$$$\left.100x^2+300x-559=0\quad\right|\colon100$$$$\left.x^2+3x-\frac{559}{100}=0\quad\right|\text{pq-Formel}$$$$x_{1;2}=-\frac32\pm\sqrt{\left(-\frac32\right)^2+\frac{559}{100}}=-\frac32\pm\sqrt{\frac94+\frac{559}{100}}=-\frac32\pm\sqrt{\frac{225}{100}+\frac{559}{100}}$$$$\phantom{x_{1;2}}=-\frac{3}{2}\pm\sqrt{\frac{784}{100}}=-\frac32\pm\frac{28}{10}=-\frac{15}{10}\pm\frac{28}{10}$$Damit haben wir die beiden Lösungen:$$x_1=\frac{13}{10}=1,3\quad;\quad x_2=-\frac{43}{10}=-4,3$$