How would you solve this with the quadratic formula and leave the answer in

simplest radical form?

y² + 6y - 1 = 0

simplest radical form?

y² + 6y - 1 = 0

Quadratic formula

-b ±√b² - 4ac

------------------ = x

2a

y² + 6y - 1 = 0

a = 1, b = 6, c = -1

-6 ± √6² - 4(1)(-1)

---------------------- = y {substituted into the quadratic formula}

2(1)

-6 ± √36 + 4

--------------- = y {squared the 6 and multiplied on the end}

2

-6 ± √40

--------- = y {added 36 and 4}

2

-6 ± √4√10

------------- = y {broke 40 into 4 and 10}

2

-6 ± 2√10

--------- = y {evaluated square root of 4 to be 2}

2

2(-3 ± √10)

--------------- {factored 2 out of top}

2

**y = -3 ± √10**{cancelled the 2's}

*© Algebra House*