# Find the discriminant of this quadratic equation then state the number and type of solution. 2x^2+4x-8= -10

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For the quadratic equation `ax^2+bx+c=0` , the discriminant is defined

`D=b^2-4ac` . Depending on the values of `D` , foolowing case arise.

(1) If `D>0, ` the roots are real and distinct.

(2) If `D=0,` the roots are real and equal.

(3) If `D<0,` the roots are imaginary.

Given equation is `2x^2+4x-8=-10`

or, `2x^2+4x-8+10=0`

or, `2x^2+4x+2=0` .

Here we see `a=2,b=4,c=2`

Now, `D=4^2-4.2.2=16-16=0` .

From case (2) we observe that given equation has real and equal roots. Now we solve the given equation.

`2x^2+4x+2=0`

```rArr` `2x^2+2x+2x+2=0`

`rArr` `2x(x+1)+2(x+1)=0`

```rArr` `(2x+2)(x+1)=0`

So, `2x+2=0 ` `rArr` `2x=-2`

`rArr` `x=-1.`

Now `x+1=0` `rArr` `x=-1` .

So we see that both the roots of the given quadratic equation are real and equal i.e. `-1,-1.`

`2x^2+4x-8=-10`

`2x^2+4x-8+10=0`

`2x^2+4x+2=0`

`a=2,b=4,c=2`

`D=b^2-4ac`

`=4^2-4xx2xx2`

`=16-16`

`=0`

since D=0

therefore roots are real and equal.

factor out 2,

`2(x^2+2x+1)=0`

`x^2+2x+1=0`

`(x+1)^2=0`

`x+1=0`

`x=-1`

Thus roots are x=-1,-1

`2x^2+4x-8= -10 ` add 10

`2x^2+4x+2=0`

`a= 2` ` b= 4 ` `c=2 ` use the formula` b^2-4ac`

`4^2-4(2)(2) ` simplify it

`16-16=0 ` **0 is the discriminant** meaning there is **1 real solution**

to find the roots use the quadratic equation

`(-4+-sqrt(0))/(2(2))`

`(-4+-0)/4 = (-4)/4 = -1`

**-1 is the root**