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What is the solution set of x^{2} +5x-5=0?
5 months ago
Q:
What is the solution set of x^{2} +5x-5=0?
Accepted Solution
A:
We can find the solution using the quadratic formula:
[tex]x= \frac{-b+- \sqrt{ b^{2} -4ac} }{2a} [/tex]
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:
[tex]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} [/tex]