a2+b2+c2-2ab+2bc-2ca
If you mean (a-b+c)^2, then... a^2 - ab + ac - ab + b^2 - bc + ac - bc + c^2 = a^2 + b^2 + c^2 - 2ab + 2ac - 2bc.
a - b = c -(a - b) = -c b - a = -c
complement of c
X2+5x-6. a=1, b=5, and c=-6 The formula is: -b plus or minus the square root of b squared minus 4ac all over 2a. -b+square root of b2-4ac ---- 2a -5 plus or minus the square root of 5 squared minus 4(1)(-6) -5 plus or minus the square root of 25-4(-6) -5 plus or minus the square root of 25+24 -5 plus or minus the square root of 49 -5 plus or minus 7 Here is where you split into two different answers: Number 1: -5 plus 7= 2 Number 2: -5 minus 7= -12 Your answer is X=2, -12
(b-c)(a+b)-ac
If you mean (a-b+c)^2, then... a^2 - ab + ac - ab + b^2 - bc + ac - bc + c^2 = a^2 + b^2 + c^2 - 2ab + 2ac - 2bc.
(a + b + c)3 = a3 + b3 + c3 + 3(a2b + ab2 + b2c + bc2 + ca2 + ac2) + 6abc
a - b = c -(a - b) = -c b - a = -c
(a+b+c)²=a²+b²+c²+ 2ab+2bc+2ac
B squared equals c squared minus a squared then to find B take the square root of you answer for b squared
its B
The expression ( (A - b + C)^2 ) can be expanded using the formula for the square of a binomial, which states that ( (x + y)^2 = x^2 + 2xy + y^2 ). In this case, it expands to ( (A - b + C)(A - b + C) ), resulting in ( A^2 - 2Ab + 2AC + b^2 - 2bC + C^2 ). Thus, the value of ( (A - b + C)^2 ) is ( A^2 + b^2 + C^2 - 2Ab + 2AC - 2bC ).
If your Problem is organized like this: A x squared plus B x plus C, the equation is: (B plus or minus the square root of(B squared minus 4 A C)) over 2A
1.774225a
(a+b-c)2 = a2 + b2 +c2 +2ab - 2bc - 2ac
the square root of b squared minus 4 times a times c
complement of c