# Find the discriminant of this quadratic equation then state the number and type of solutions. 6b^2+b+3= -5

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Because given equation is a quadratic equation i.e. of degree 2. So, this equation will always have two solutions (either real or complex).

The discriminant for the quadratic equation `px^2+qx+r=0`

is given by `D=q^2-4pr` .

Now if (i) `D>0` , both roots are real and distinct

(ii) `D=0` , both roots are real and equal

(iii) `D<0` , both roots are complex.

Given equation is `6b^2+b+3=-5`

or, `6b^2+b+3+5=0`

or, `6b^2+b+8=0`

Here `p=6, q=1, r=8.`

So, `D=1^2-4.6.8=1-192=-191<0` .

So, by condition (iii) both roots of the given quadratic equation will be complex.

`6b^2+b+3 = -5`

`6b^2+b+3+5 = 0`

`6b^2+b+8 = 0`

Discriminant = `(1)^2-4xx6xx8 = -191 < 0`

Since the discriminant is negative there are no real solutions to the above function. But there are two imaginary solutions for this using complex numbers.

`6b^2+b+3=-5`

adding 5 both sides: `6b^2+b+3+5= -5+5` `6b^2+b+8=0```

`Delta=1^2-4(8)(6)= 1-192= -191`

since `Delta <0` the equation has two complex coniugates solutions:

`b=(-1 +- sqrt(Delta))/12` `=(-1 +- i sqrt(191))/12`

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`6b^2+b+3=-5`

`6b^2+b+3+5=0`

`6b^2+b+8=0`

`6(b^2+(1/6)b+(8/6))=0`

`b^2+(1/6)b+8/6=0`

`b^2+2(1/12)b+1/144=1/144-8/6`

`(b+1/12)^2=(1-192)/144`

`(b+1/12)^2=(-191)/144`

`` Discriminant=-191 <0

**so no real roots.Roots are complex conjugate**.