Question

How do I solve 3.2x+0.2x^2-5=0

Answer 1

`3.2x+0.2x^2-5=0\ \ \ \ |both\ sides\ multiply\ by\ 5\\\\16x+x^2-25=0\\\\x^2+16x-25=0\\\\a=1;\ b=16;\ c=-25\\\\\Delta=b^2-4ac\to\Delta=16^2-4\cdot1\cdot(-25)=256+100=356\\\\x_1=(-b-\sqrt\Delta)/(2a)\ and\ x_2=(-b+\sqrt\Delta)/(2a)\\\\\sqrt\Delta=√(356)=√(4\cdot89)=2√(89)\\\\x_1=(-16-2√(89))/(2\cdot1)=-8-√(89);\ x_2=(-16+2√(89))/(2\cdot1)=-8+√(89)`