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

rakesh05 | Certified Educator

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.`

lemjay | Certified Educator

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).

oldnick | Student

`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.

atyourservice | Student

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

pramodpandey | Student

`-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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