Question

What is the solution set of x^{2} +5x-5=0?

Answer 1

We can find the solution using the quadratic formula:

$x= \frac{-b+- \sqrt{ b^{2} -4ac} }{2a}$

b =coefficient of x term = 5
a = coefficient of squared term = 1
c = constant term = -5

Using the values in the above formula we get:

$x= \frac{-5+- \sqrt{25-4(1)(-5)} }{2(1)} \\ \\ x= \frac{-5+- \sqrt{45} }{2} \\ \\ x= \frac{-5+-3 \sqrt{5} }{2} \\ \\ x= \frac{-5+3 \sqrt{5} }{2}, x= \frac{-5-3 \sqrt{5} }{2}$

Answer 2

Answer:We can find the solution using the quadratic formula:

x= \frac{-b+- \sqrt{ b^{2} -4ac} }{2a}  

b =coefficient of x term = 5

a = coefficient of squared term = 1

c = constant term = -5

Using the values in the above formula we get:

x= \frac{-5+- \sqrt{25-4(1)(-5)} }{2(1)} \\  \\ 
x= \frac{-5+- \sqrt{45} }{2} \\  \\ 
x= \frac{-5+-3 \sqrt{5} }{2} \\  \\ 
x= \frac{-5+3 \sqrt{5} }{2}, x= \frac{-5-3 \sqrt{5} }{2}  

Step-by-step explanation: