Question

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

Answer 1

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}

Explanation:

Answer 2

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= (-5+- √(25-4(1)(-5)) )/(2(1)) \\ \\ x= (-5+- √(45) )/(2) \\ \\ x= (-5+-3 √(5) )/(2) \\ \\ x= (-5+3 √(5) )/(2), x= (-5-3 √(5) )/(2)`