# Find the discriminant of this quadratic equation then state the number and type of solution. `-5x^2+x-19= -9`

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### 5 Answers

The formula of the discriminant of a quadratic equation in a form ax^2+bx+c=0 is b^2-4ac.

Before we apply the formula for the given equation, set one side equal to zero. To do so, add both sides by 9.

`-5x^2+x-19+9=-9+9`

`-5x^2+x-10=0`

Then, plug-in the value of a b and c to the formula of discriminant . The values are a=-5, b=1 and c=-10.

`b^2-4ac=1^2-4(-5)(-10)=1-200=-199`

**Hence, the discriminant of the given equation is -199.**

**Since its value is less than zero, the quadratic equation has two complex solutions (non-real numbers).**

Given quadratic equation is `-5x^2+x-19=-9` . Which can be written as `-5x^2+x-19+9=0`

or, `-5x^2+x-10=0.`

The discriminant for the quadratic equation `ax^2+bx+c=0` is defined as `D=b^2-4ac.` Since the given equation is quadratic, it has two solutions described as under-

i) If `D>0` , both roots of the equation are real and distinct.

ii) If `D=0` , both roots of the equation are real and equal.

iii) If `D<0` , both roots of the equation are complex.

Here we see that `a=-5,b=1,c=-10` .

Now, `D=1^2-4(-5)(-10)=1-4.50=1-200=-199.`

From case (iii) it is clear that roots of given quadratic equation are complex i.e. of the form `a+-ib` .i.e. the first root will be of the form `a+ib` and the second root will be of the form `a-ib.`

`5x^2-x+10=0` the a =5 b= -1 c= 10 so:

`Delta= b^2-4ac= (-1)^2- 4( 10)(5)= -199`

since negative, hsn's realsolution but two comple solution.

add the 9 to the other side

`-5x^2+x-19+9=-9+9 ` you will end up with

`-5x^2+x-10=0`

`a=-5 b=1 c=-10` the formula for finding the discriminant is b^2-4ac plug in the numbers into the formula

`1^2-4(-5)(-10)`

`1-200=-199` the discriminant is -399 since it is less than 0 it means there are **no real solutions but instead two complex solutions**

Given quadratic equation

`-5x^2+x-19=-9`

`-5x^2+x-19+9=0`

`-5x^2+x-10=0`

`-5(x^2-(1/5)x+2)=0`

`x^2-2(1/10)x+(1/10)^2=-2+(1/10)^2`

`(x-1/10)^2=-199/100`

`(x-1/10)^2=(199i^2)/100 ,i^2=-1`

`x-1/10=+-(isqrt(199))/10`

`x=(1+-isqrt(199))/10`

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