# Find the constant k so that the quadratic equation 2x^2 + 5x - k = 0 has two real solutions

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Given the quadratic equation f(x) = 2x^2 + 5x -k

We need to find the value of k such that f(x) has two real solutions.

We will use delta to determine the values of k.

We know that the function has two real solution is delta > 0

delta = b^2 - 4ac

We know that: a = 2 b= 5 c = -k

==> b^2 - 4ac = (25 - 4*2*-k ) > 0

==> 25 + 8k > 0

==> 8k > -25

==> k > - 25/8

**Then the values of k belongs to the interval ( -25/8 , inf)**

We have to find the value of k so that the quadratic equation 2x^2 + 5x - k = 0 has two real solutions.

Now the roots of a quadratic equation are given by [-b+ sqrt(b^2 - 4ac]/2a and [-b- sqrt(b^2 - 4ac]/2a.

For the roots to be real and different b^2 - 4ac > 0

=> 5^2 - 4*-k*2 > 0

=> 25 + 8k > 0

=> 8k > -25

=> 8k > -25/8

=> k > -25/8

**Therefore k should be greater than -25/8.**