Question

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

Answer 1

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