Answer:
The solution to the equation are
![5+(√(42) )/(2\\) \ and \ 5-(√(42) )/(2\\)\\](https://img.qammunity.org/2021/formulas/mathematics/high-school/goqynzki0ln22lr3xi064x3kx6956awswp.png)
Both of his values are positive real numbers
Explanation:
The general formula of a quadratic equation is expressed as
![ax^(2)+bx+c = 0\ where;\\x = -b\±\frac{\sqrt{b^(2)-4ac } }{2a}](https://img.qammunity.org/2021/formulas/mathematics/high-school/ubnxjs4bxwn9rlpewrbtytt8ey8ylikc1o.png)
Given the expression 0 = x² – 5x – 4 which can be rewritten as shown below;
x² – 5x – 4 = 0
Comparing this to the general equation; a = 1, b = -5, c= -4
To get the solution to the quadratic equation, we will use the general formula above;
![x = -b\±\frac{\sqrt{b^(2)-4ac } }{2a}\\x = -(-5)\±\frac{\sqrt{(-5)^(2)-4(1)(-4) } }{2(1)}\\\\x = 5\±(√(25+16 ) )/(2)\\x =5\±(√(41) )/(2)\\x = 5+(√(42) )/(2)\ and \ 5-√(42) /2\\](https://img.qammunity.org/2021/formulas/mathematics/high-school/8vwpmihwn4skwzzgrxb5rk1t04bbc9a8s7.png)
Both of his values are positive real numbers