Question

In the quadratic formula is used to find the solution set of 3x^2+4x-2=0

Answer 1

`ax^2+bx+c=0\\\\\Delta=b^2-4ac\\\\if\ \Delta > 0\ then\ two\ solutions\\\\x_1=(-b-\sqrt\Delta)/(2a);\ (-b+\sqrt\Delta)/(2a)\\\\if\ \Delta=0\ the\ one\ solution\\\\x_o=(-b)/(2a)\\\\if\ \Delta < 0\ then\ no\ solutions.`
We have:
`3x^2+4x-2=0\\\\a=3;\ b=4;\ c=-2`
sustitute:
`\Delta=4^2-4\cdot3\cdot(-2)=16+24=40\\\\\sqrt\Delta=√(40)=√(4\cdot10)=\sqrt4\cdot√(10)=2√(10) > 0`

`x_1=(-4-2√(10))/(2\cdot3)=(-2-√(10))/(3)\\\\x_2=(-4+2√(10))/(2\cdot3)=(-2+√(10))/(3)`