# Order of Operations: -6(u+5)^2=120 Should this be divided by 6 first or should it be squared first?

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You can operate both way because it works. Here it is the third way to solve the problem, hence, moving 120 to the left yields:

`-6(u+5)^2 - 120 = 0`

Factoring out -6 yields:

`-6((u + 5)^2 + 20) = 0 => (u + 5)^2 + 20 = 0 => u^2 + 10u + 25 + 20 = 0 => u^2 + 10u + 45 = 0`

You may use uadratic formula such that:

`u_(1,2) = (-10+-sqrt(10^2 - 4*1*45))/2 => u_(1,2) = (-10+-sqrt-80)/2 => u_(1,2) = (-10 +- 4*i*sqrt5)/2`

`u_(1,2) = -5 +- 2*isqrt5`

Hence, evaluating the solutions to quadratic equation, using complex number theory, yields `u_(1,2) = -5 +- 2*isqrt5` .

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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Let's think a bit more.

Let's suppose that we'll divide by -6, both sides, first:

(u+5)^2 = -20

We've get an impossible situation for real numbers: something raised to square is always positive.

This situation could be solved if we would work in the imaginary field.

Let's apply square root both sides:

u + 5 = sqrt(-20)

In imaginary field, sqrt -1 = i

u + 5 = 2isqrt5

u = -5 + 2isqrt5

u + 5 = -2isqrt5

u = -5 -2isqrt5

Let's come back to the first situation. We are in the real numbers field and the situation (u+5)^2 = -20 is impossible.

That leads us to another approach.

We'll not divide by -6 first, but we'll raise to square first:

-6(u^2 + 10u + 25) = 120

We'll remove the brackets:

-6u^2 - 60u - 150 - 120 = 0

We'll multiply by -1:

6u^2 + 60u + 270 = 0

We'll divide by 2:

3u^2 + 30u + 135 = 0

We'll divide by 3:

u^2 + 10u + 45 = 0

u1 = [-10+sqrt(100-180)]/2

We can see that the discriminant of the equation is negative, so the equation has no real roots.

In other words, no real value of u will make the equality to hold.