# Use the discriminant to predict the nature of the roots, then use the quadratic formula to find the roots. 6x^2 + 7x - 3 = 0

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The nature of the roots of a quadratic equation ax^2 + bx + c = 0 can be determined from the discriminant D = b^2 - 4ac. If D > 0, the roots are real and unique, if D = 0, the equation has equal real roots and if D < 0, the roots are complex.

For 6x^2 + 7x - 3 = 0, D = 7^2 -(4*6*-3) = 121

The roots are given by `(-b+-sqrt D)/(2*a)`

= `(-7+-11)/12`

**The roots of the equation 6x^2 + 7x - 3 = 0 are **`(-7+-11)/12`

`6x^2 + 7x - 3 = 0 ` use the formula `b^2-4ac`

`a=6 ` `b=7 ` `c=-3 ` plug it in

`7^2-4(6)(-3)` simplify it

`49+72=121` the discriminant is 121, meaning the problem has 2 real solutions as 121 is bigger than 0.

to find the roots use the quadratic formula `(-b+-sqrt(b^2-4ac))/(2a)`

`(-7+-sqrt(121))/(2(6))`

`(-7+-sqrt(121))/(12)`

`(-7+-11)/(12)`

`6x^2 + 7x - 3 = 0`

Discriminant= `7^2-4*6*(-3)=121`

Since, the value of the discriminant is positive and a perfect square, the given quadratic equation has two real, rational roots.

Using the quadratic formula the roots are:

`(-7+-sqrt(7^2-4*6*(-3)))/(2*6)`

`=(-7+-11)/12`

`=-3/2,1/3`

Hence, the roots of the given equation are `-3/2` and `1/3` .

The graph: