How do you solve #30 + x − x^2 = 0#?

3 Answers
Aug 1, 2015

Answer:

#x=-5,6#

Explanation:

Invert (multiply by -1, has the same solutions) and complete the square:
#x^2-x-30=(x-1/2)^2-121/4=0#

Solve for #x#:
#(x-1/2)^2=121/4#

=>

#x-1/2=+-11/2#

=>

#x=(1+-11)/2#

Aug 1, 2015

Answer:

solve #y = -x^2 + x + 30 = 0#

Ans: -5 and 6

Explanation:

I use the new Transforming Method (Google, Yahoo, Bing Search)
Find 2 numbers knowing sum (1) and product (-30). Roots have opposite signs since a and c have opposite signs.
Factor pairs of (-30) --> (-2, 15)(-4, 5)(-5, 6). This sum is 1 = b.
Since a < 0. then the 2 real roots are: -5 and 6.

Aug 1, 2015

Answer:

You could use the quadratic formula.

Explanation:

First, rewrite your quadratic in the form

#color(blue)(ax^2 + bx + c = 0)#

for which the quadratic formula takes the form

#color(blue)(x_(1,2) = (-b +- sqrt(b^2 - 4ac))/(2a)#

You will start from

#-x^2 + x + 30= 0#

which can be rewritten as

#-(x^2 - x - 30) = 0#

In this case, #a=11#, #b=-1#, and #c=-30#.

The two solutions to this quadratic equation will thus be

#x_(1,2) = (-(-1) +- sqrt( (-1)^2 - 4 * (1) * (-30)))/(2 * (1))#

#x_(1,2) = (1 +- sqrt(121))/(-2) = (1 +-11)/2#

#x_1 = (1 + 11)/(2) = color(green)(6)#

#x_2 = (1 - 11)/(2) = color(green)(-5)#