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