Solve using the Square Root Property (y-5)(y-2)=28

Math
(y-5)(y-2)=28
Simplify (y-5)(y-2).
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Expand (y-5)(y-2) using the FOIL Method.
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Apply the distributive property.
y(y-2)-5(y-2)=28
Apply the distributive property.
y⋅y+y⋅-2-5(y-2)=28
Apply the distributive property.
y⋅y+y⋅-2-5y-5⋅-2=28
y⋅y+y⋅-2-5y-5⋅-2=28
Simplify and combine like terms.
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Simplify each term.
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Multiply y by y.
y2+y⋅-2-5y-5⋅-2=28
Move -2 to the left of y.
y2-2⋅y-5y-5⋅-2=28
Multiply -5 by -2.
y2-2y-5y+10=28
y2-2y-5y+10=28
Subtract 5y from -2y.
y2-7y+10=28
y2-7y+10=28
y2-7y+10=28
Move 28 to the left side of the equation by subtracting it from both sides.
y2-7y+10-28=0
Subtract 28 from 10.
y2-7y-18=0
Factor y2-7y-18 using the AC method.
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Consider the form x2+bx+c. Find a pair of integers whose product is c and whose sum is b. In this case, whose product is -18 and whose sum is -7.
-9,2
Write the factored form using these integers.
(y-9)(y+2)=0
(y-9)(y+2)=0
If any individual factor on the left side of the equation is equal to 0, the entire expression will be equal to 0.
y-9=0
y+2=0
Set the first factor equal to 0 and solve.
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Set the first factor equal to 0.
y-9=0
Add 9 to both sides of the equation.
y=9
y=9
Set the next factor equal to 0 and solve.
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Set the next factor equal to 0.
y+2=0
Subtract 2 from both sides of the equation.
y=-2
y=-2
The final solution is all the values that make (y-9)(y+2)=0 true.
y=9,-2
Solve using the Square Root Property (y-5)(y-2)=28

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