Solve for y 2y^2+24y+12=y^2+32-3

Math
2y2+24y+12=y2+32-3
Subtract 3 from 32.
2y2+24y+12=y2+29
Move all terms containing y to the left side of the equation.
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Subtract y2 from both sides of the equation.
2y2+24y+12-y2=29
Subtract y2 from 2y2.
y2+24y+12=29
y2+24y+12=29
Move all terms to the left side of the equation and simplify.
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Move 29 to the left side of the equation by subtracting it from both sides.
y2+24y+12-29=0
Subtract 29 from 12.
y2+24y-17=0
y2+24y-17=0
Use the quadratic formula to find the solutions.
-b±b2-4(ac)2a
Substitute the values a=1, b=24, and c=-17 into the quadratic formula and solve for y.
-24±242-4⋅(1⋅-17)2⋅1
Simplify.
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Simplify the numerator.
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Raise 24 to the power of 2.
y=-24±576-4⋅(1⋅-17)2⋅1
Multiply -17 by 1.
y=-24±576-4⋅-172⋅1
Multiply -4 by -17.
y=-24±576+682⋅1
Add 576 and 68.
y=-24±6442⋅1
Rewrite 644 as 22⋅161.
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Factor 4 out of 644.
y=-24±4(161)2⋅1
Rewrite 4 as 22.
y=-24±22⋅1612⋅1
y=-24±22⋅1612⋅1
Pull terms out from under the radical.
y=-24±21612⋅1
y=-24±21612⋅1
Multiply 2 by 1.
y=-24±21612
Simplify -24±21612.
y=-12±161
y=-12±161
The final answer is the combination of both solutions.
y=-12+161,-12-161
The result can be shown in multiple forms.
Exact Form:
y=-12+161,-12-161
Decimal Form:
y=0.68857754…,-24.68857754…
Solve for y 2y^2+24y+12=y^2+32-3

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