+130 I need help solving the following:
35-b^2=2b
Which is shortened alot already, but I can't seem to find a way to continue on the equation?
The Web.2.0calc.com calculator only gives me the answer, which is not intresting. I need to know HOW to solve this!
Thanks, Best Regars, Mormert.
+36916 ....or you can just factor it:
35-b^2 = 2b add 'b^2 - 35 ' to both sides
0=b^2 + 2b - 35
(b+7)(b-5) = 0 Now either b+7 or b-5 (or both) = 0 so b = -7 or 5
+14915 Hello, good morning Mormert!
35-b^2=2b b = ?
The general form of a quadratic equation is
ax² + bx + c = 0
x is calculated using the equation solution
\({\color{Blue}x = {-b \pm \sqrt{b^2-4ac} \over 2a}}\)
Put the equation in the general form.
35-b²=2b
-b² - 2b + 35 = 0 Replacing b = x
-x² - 2x + 35 = 0
ax² + bx + c = 0
a=-1 b=-2 c=35
\({\color{Blue}x = {-b \pm \sqrt{b^2-4ac} \over 2a}}\)
\({\color{Blue}x = {-(-2) \pm \sqrt{(-2)^2-4*(-1)*35} \over 2*(-1)}}\)
\({\color{Blue}x=\frac{2\pm \sqrt{144} }{-2} = \frac {2\pm12}{-2}}\)
x1 = - 7
x2 = 5
Swap back x b
b1 = - 7
b2 = 5 
!
I hope I could help you.
asinus: -)!
+36916 ....or you can just factor it:
35-b^2 = 2b add 'b^2 - 35 ' to both sides
0=b^2 + 2b - 35
(b+7)(b-5) = 0 Now either b+7 or b-5 (or both) = 0 so b = -7 or 5
