Solve for b 48=b^2-2b

Math
48=b2-2b
Rewrite the equation as b2-2b=48.
b2-2b=48
Move 48 to the left side of the equation by subtracting it from both sides.
b2-2b-48=0
Factor b2-2b-48 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 -48 and whose sum is -2.
-8,6
Write the factored form using these integers.
(b-8)(b+6)=0
(b-8)(b+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.
b-8=0
b+6=0
Set the first factor equal to 0 and solve.
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Set the first factor equal to 0.
b-8=0
Add 8 to both sides of the equation.
b=8
b=8
Set the next factor equal to 0 and solve.
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Set the next factor equal to 0.
b+6=0
Subtract 6 from both sides of the equation.
b=-6
b=-6
The final solution is all the values that make (b-8)(b+6)=0 true.
b=8,-6
Solve for b 48=b^2-2b

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