\(3x^2+7x-20=0\\\\a=3;\ b=7;\ c=-20\\\\\Delta=b^2-4ac\\\\x_1=\frac{-b-\sqrt\Delta}{2a};\ x_2=\frac{-b+\sqrt\Delta}{2a}\\\\\Delta=7^2-4\cdot3\cdot(-20)=49+240=289;\ \sqrt\Delta=\sqrt{289}=17\\\\x_1=\frac{-7-17}{2\cdot3}=\frac{-24}{6}=-4;\ x_2=\frac{-7+17}{2\cdot3}=\frac{10}{6}=\frac{5}{3}\\\\\\3x^2+7x-20=3(x+4)(x-\frac{5}{3})\\\\\\3x^2+7x-20=0\iff x=-4\ \vee\ x=\frac{5}{3}\)