How do you solve $root3(6u-5)+2=-3$ ?

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Answer 1 · 2017-01-15T18:13:40

Move the 2 to the right.
Cube both sides.
Then solve for u.

Given: $root(3)(6u - 5) + 2 = -3$

Subtract 2 from both sides:

$root(3)(6u - 5) = -5$

Cube both sides:

$6u - 5 = -125$

Add 5 to both sides:

$6u = -120$

Divide both sides by 6:

$u = -20$

check:

$root(3)(6(-20) - 5) + 2 = -3$

$root(3)(-120 - 5) + 2 = -3$

$root(3)(-125) + 2 = -3$

$-5 + 2 = -3$

$-3 = -3$

This checks.

How do you solve root3(6u-5)+2=-3root3(6u-5)+2=-3?