Question

Determine the number of solutions to the quadratic equation: 5z^(2)+6z-2=0

Answer 1

Derivation of the quadratic formula:

`ax^2+bx+c=0\\\mathrm{or,\ }ax^2+bx=-c\\\mathrm{or,\ }4a^2x^2+4abx=-4ac\\\mathrm{or,\ }4a^2x^2+4abx=b^2-4ac\\\mathrm{or,\ }(2ax+b)^2=b^2-4ac\\\mathrm{or,\ }2ax+b=\pm√(b^2-4ac)\\\therefore\ x=(-b\pm√(b^2-4ac))/(2a)`

Answer:

`5z^2+6z-2=0\\\mathrm{or,\ }z=(-6\pm√(6^2-4(5)(-2)))/(2(5))=(-6\pm√(76))/(10)=(-6\pm2√(19))/(10)=(-3\pm√(19))/(5)`

`\mathrm{Therefore\ the\ quadratic\ equation\ has\ two\ solutions,\ (-3+√(19))/(5)\ and\ (-3-√(19))/(5).}`

Answer 2

two solutions

Explanation:

the number of solutions to this quadratic equation are two.

because a quadratic has always two solutions