# Find the discriminant of this quadratic equation then state the number and type of solution. `-3x^2-7x+8=10`

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We know that the discriminant for the quadratic equation `ax^2+bx+c=0` is `D=b^2-4ac.`

Now we have the following conditions based on the values of `D` :

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

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

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

Now given problem is `-3x^2-7x+8=10` . Which can be written as

`-3x^2-7x+8-10=0`

or, `-3x^2-7x-2=0`

Here `a=-3, b=-7, c=-2.`

So, `D=(-7)^2-4(-3)(-2)=49-24=25>0`

So, by condition (i) both roots of the given quadratic equation are real and distinct.

Our equation can be written as

`3x^2+7x+2=0` , multiplying both sides by (-1).

so, `3x^2+6x+x+2=0`

or, `3x(x+2)+1(x+2)=0`

or, `(3x+1)(x+2)=0`

Now `(3x+1)=0rArr x=-1/3` ,

and `(x+2)=0rArr x=-2` .

So, `x=-1/3, -2.`

The discriminant of quadratic equation ax^2 + bx +c = 0 is b^2-4ac.

But before applying this formula, set one side of the given equation equal to zero. To do so, subtract both sides by 10.

`-3x^2-7x+8=10`

`-3x^2-7x+8-10=10-10`

`-3x^2-7x-2=0`

Then, plug-in the values of a, b and c to the formula of discriminant. The values are a=-3, b=-7 and c=-2.

`b^2-4ac=(-7)^2-4(-3)(-2)=49-24=25` **Hence, the discriminant is 25.****Also, since the value of discriminant is greater tha zero and a perfect square (25=5^2), therefore, the solutions of the given equation are two rational numbers.**

take away 10

`-3x^2-7x+8-10`

`-3x^2-7x-2`

`a=-3` `b=-7` `c=-2` use `b^2-4ac`

`-7^2-4(-3)(-2)`

`49-24=25 ` the discriminant is 25 and since it is bigger than 0 it means it has 2 real solutions. The roots can be found using the quadratic equation.

`(7+-sqrt(25))/(2(-3))`

`(7+-5)/(-6)`

`(7+5)/(-6) = (12)/(-6) = -2`

`(7-5)/(-6) = 2/(-6) = 1/-3`

the roots are -2 and 1/-3