# Please determine the number of solutions of x^2 - 5x - 7 = 0.  Please show work. You may use quadratic formula to evaluate solutions to the given quadratic equation such that:

`x_(1,2) = (-b+-sqrt(b^2-4ac))/(2a)`

You need to identify the coefficients a,b,c such that:

`a = 1 , b = -5 , c = -7`

Substituting the above values for a,b,c yields:

`x_(1,2) = (-(-5)+-sqrt(25+28))/2`

`x_(1,2) = (5+-sqrt(53))/2`

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You may use quadratic formula to evaluate solutions to the given quadratic equation such that:

`x_(1,2) = (-b+-sqrt(b^2-4ac))/(2a)`

You need to identify the coefficients a,b,c such that:

`a = 1 , b = -5 , c = -7`

Substituting the above values for a,b,c yields:

`x_(1,2) = (-(-5)+-sqrt(25+28))/2`

`x_(1,2) = (5+-sqrt(53))/2`

You may also use the alternate method, hence, you should complete the square using the formula `(a-b)^2 = a^2 - 2ab + b^2`  such that:

`x^2 -5x = x^2 - 2*(5/2)x + (5/2)^2 - (5/2)^2`

`x^2 - 5x = (x - 5/2)^2 - 25/4`

`x^2 - 5x - 7 = 0 =>(x - 5/2)^2 - 25/4 - 7 =`  0

You need to keep the squared binomial to the left side such that:

`(x - 5/2)^2= 25/4 + 7`

`(x - 5/2)^2= (25+28)/4 => x - 5/2 = +-sqrt(54/4)`

`x_(1,2) = 5/2+-(sqrt53)/2`

Hence, evaluating the solutions to the given quadratic equation, using both methods, yields `x_(1,2) = (5+-sqrt(53))/2.`

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