b2+5=8b−10

Question

b2+5=8b−10

Gennenzip · Accepted Answer

To solve a quadratic equation, we need to first move all the terms to one side.
Subtracting 8b on both sides of $b 2 + 5 = 8 + 5 = 8 b - 10$ gives $b 2 - 8 b + 5 = - 10 .$
Adding 10 on both sides then gives $b 2 - 8 b + 15 = 0 .$
Now that we have all the terms on one side, we want to see if the quadratic is factorable. If it's not factorable, we will need to use the Quadratic Formula to find the solutions.
A quadratic of the form $x^{2} + B x + C x$ can be factored to $(x + m) (x + n)$ if there are two numbers, m and n, such that m+n=B and mn=C.
For $b 2 - 8 b + 15$ , we then need to see if there are two numbers that multiply to 15 and add to −8. Since $- 5 (- 3) = 15$ and $- 5 + (- 3) = - 8$ , then the two numbers will be −5 and −3.
$b^{2} - 8 b + 15 b$ can then factor to (b−5)(b−3).
The equation $b^{2} - 8 b + 15 = 0$ can then be written as (b−5)(b−3)=0.
The Zero Product Property states that if ab=0, then a=0 or b=0. Using this property, if $(b - 5) (b - 3) = 0$ , we can then say that b−5=0 or b−3=0.
Solving b−5=0 for bb gives b=5. Solving b−3=0 for b gives b=3.
The two solutions of the equation are then b=3 and b=5.

b^{2}+5=8b−10

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