Solve for y 2y^2+13y+10=(y+4)^2

Math
2y2+13y+10=(y+4)2
Simplify (y+4)2.
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Rewrite (y+4)2 as (y+4)(y+4).
2y2+13y+10=(y+4)(y+4)
Expand (y+4)(y+4) using the FOIL Method.
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Apply the distributive property.
2y2+13y+10=y(y+4)+4(y+4)
Apply the distributive property.
2y2+13y+10=y⋅y+y⋅4+4(y+4)
Apply the distributive property.
2y2+13y+10=y⋅y+y⋅4+4y+4⋅4
2y2+13y+10=y⋅y+y⋅4+4y+4⋅4
Simplify and combine like terms.
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Simplify each term.
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Multiply y by y.
2y2+13y+10=y2+y⋅4+4y+4⋅4
Move 4 to the left of y.
2y2+13y+10=y2+4⋅y+4y+4⋅4
Multiply 4 by 4.
2y2+13y+10=y2+4y+4y+16
2y2+13y+10=y2+4y+4y+16
Add 4y and 4y.
2y2+13y+10=y2+8y+16
2y2+13y+10=y2+8y+16
2y2+13y+10=y2+8y+16
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+13y+10-y2=8y+16
Subtract 8y from both sides of the equation.
2y2+13y+10-y2-8y=16
Subtract y2 from 2y2.
y2+13y+10-8y=16
Subtract 8y from 13y.
y2+5y+10=16
y2+5y+10=16
Move 16 to the left side of the equation by subtracting it from both sides.
y2+5y+10-16=0
Subtract 16 from 10.
y2+5y-6=0
Factor y2+5y-6 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 -6 and whose sum is 5.
-1,6
Write the factored form using these integers.
(y-1)(y+6)=0
(y-1)(y+6)=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-1=0
y+6=0
Set the first factor equal to 0 and solve.
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Set the first factor equal to 0.
y-1=0
Add 1 to both sides of the equation.
y=1
y=1
Set the next factor equal to 0 and solve.
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Set the next factor equal to 0.
y+6=0
Subtract 6 from both sides of the equation.
y=-6
y=-6
The final solution is all the values that make (y-1)(y+6)=0 true.
y=1,-6
Solve for y 2y^2+13y+10=(y+4)^2

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