What are the solutions to the quadratic equation (5y + 6)2 = 24? y = StartFraction negative 6 + 2 StartRoot 6 EndRoot Over 5 End

Question

2 months ago

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What are the solutions to the quadratic equation (5y + 6)2 = 24? y = StartFraction negative 6 + 2 StartRoot 6 EndRoot Over 5 End

Fraction and y = StartFraction negative 6 minus 2 StartRoot 6 EndRoot Over 5 EndFraction y = StartFraction negative 6 + 2 StartRoot 6 EndRoot Over 5 EndFraction and y = StartFraction 6 minus 2 StartRoot 6 EndRoot Over 5 EndFraction y = StartFraction negative 4 StartRoot 6 EndRoot Over 5 EndFraction and y = StartFraction negative 8 StartRoot 6 EndRoot Over 5 EndFraction y = StartFraction 4 StartRoot 6 EndRoot Over 5 EndFraction and y = StartFraction 8 StartRoot 6 EndRoot Over 5 EndFraction

Mathematics

2 answers:

Leona [12.6K] · Answer 1 · 2026-04-01T21:49:40+03:00

Response:

The equation yields two distinct solutions:

a) $y=\frac{-6+2\sqrt{6} }{5}$ , and

b) $y=\frac{-6-2\sqrt{6} }{5}$

Detailed breakdown:

Given the equation: $(5y+6)^2=24$ , since the perfect square involving the variable "y" is positioned on the left side of the equality, we will proceed by taking the square root of both sides, followed by isolating the variable:

$(5y+6)^2=24\\\sqrt{(5y+6)^2} =+/-\sqrt{24} \\(5y+6)=+/-\sqrt{6*4} \\(5y+6)=+/-2\sqrt{6}\\5y=-6+/-2\sqrt{6}\\y=\frac{-6+/-2\sqrt{6} }{5}$

Consequently, there are two solutions:

a) $y=\frac{-6+2\sqrt{6} }{5}$ , and

b) $y=\frac{-6-2\sqrt{6} }{5}$

Svet_ta [12.7K] · Answer 2 · 2026-04-01T21:52:50+03:00

Svet_ta [12.7K]2 months ago

5 0

Response: A

$y=\frac{-6+2\sqrt{6} }{5} \\y=\frac{-6-2\sqrt{6} }{5}$

Detailed explanation:

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