# Decide what is t if quadratic equation has two equal roots. x^2+(2t-4)*x+t+1=0

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A quadratic equation ax^2 + bx + c = 0 has two equal roots if b^2 = 4*a*c

Here we have x^2+(2t-4)*x+t+1=0

The roots are equal if (2t - 4)^2 = 4*(t + 1)

=> 4t^2 + 16 - 16t = 4t + 4

=> 4t^2 - 20t + 12 = 0

=> t^2 - 5t + 3 = 0

t1 = 5/2 + [sqrt (25 - 12)]/2

=> 5/2 + (sqrt 13)/2

t2 = 5/2 - (sqrt 13)/2

**The value of t is 5/2 + (sqrt 13)/2 and 5/2 - (sqrt 13)/2**

For the given quadratic to have 2 equal roots, the discriminant delta is cancelling out.

delta = (2t-4)^2 - 4(t+1)

delta = 0

We'll expand the square and we'll remove the brackets in the expression of delta:

4t^2 - 16t + 16 - 4t - 4 = 0

We'll combine like terms:

4t^2 - 20t + 12 = 0

We'll divide by 4:

t^2 - 5t + 3 = 0

We'll have to determine the roots of the expression t^2 - 5t + 3 = 0

t^2 - 5t + 3 = 0

We'll apply the quadratic formula:

t1 = [5+sqrt(25 - 12)]/2

t1 = (5 + sqrt13)/2

t2 = (5 - sqrt13)/2

**The values of t, for the given quadratic to have equal roots, are: {(5 - sqrt13)/2 ; (5 + sqrt13)/2}.**