Solve the following quadratic equations :
`x^(3)-4x-3=0`

Answer

`(2+sqrt7),(2-sqrt7)` are the roots of the given quadratic equation.

Explanation

`x^(2)-4x-3=0`
Comparing it with `ax^(2)+bx+=0,a=1,b=-4,c=-3`
`b^(2)-4ac=(-4)-4(1)(-3)`
=16+12=28
`x=(-bpmsqrt(b^(2)-4ac))/(2a)`
`=(-(-4)pmsqrt28)/(2xx1)`
`=(4pmsqrt(4xx7))/(2)`
`=(2(2pmsqrt7))/(2)`
`=2pmsqrt7`
`:.x=2+sqrt7orx=2sqrt7`