(x-5)(x+19) = x^2 + 14x - 95
second degree equation (y = a*x^2 + b*x +c), must use Delta = b^2 - 4*a*c to solve it
*** Solution ***
Delta = (14)^2 - 4*(1)*(-95) = 596 > 0
two solutions:
x1 = (-b-sqrt(Delta))/(2*a)
x2 = (-b+sqrt(Delta))/(2*a)
Numeric Application :
x1 = -19
x2 = +5
(x-5)(x+19)=585
3 answers
Joe ignored the 585 on the Right Side
(x-5)(x+19)=585
x^2 + 14x - 95 = 585
x^2 + 14x + .... = 680
x^2 + 14x + 49 = 680 + 49 using completing the square because of the nice numbers
(x+7)^2 = 729
x+7 = ±27
x = 20 or x = -34
which could have been found also by factoring
(x-20)(x+34) = 0
etc.
(x-5)(x+19)=585
x^2 + 14x - 95 = 585
x^2 + 14x + .... = 680
x^2 + 14x + 49 = 680 + 49 using completing the square because of the nice numbers
(x+7)^2 = 729
x+7 = ±27
x = 20 or x = -34
which could have been found also by factoring
(x-20)(x+34) = 0
etc.
true i am sorry for that !
thanks Reiny for correcting :)
thanks Reiny for correcting :)