# How do you solve u=sqrt(-u+12)?

Aug 29, 2017

$u = - 4 \mathmr{and} u = 3$

#### Explanation:

$u = \sqrt{- u + 12}$

Square both sides

${u}^{2} = {\sqrt{- u + 12}}^{2}$

${u}^{2} = - u + 12$

Collect like terms

${u}^{2} + u - 12 = 0 \to Q u a \mathrm{dr} a t i c$

Solving the Equation we have $+ 4 , - 3$ as the quadratic factors

${u}^{2} + 4 u - 3 u - 12 = 0$

$\left({u}^{2} + 4 u\right) \left(- 3 u - 12\right) = 0$

$u \left(u + 4\right) - 3 \left(u + 4\right) = 0$

$\left(u + 4\right) \left(u - 3\right) = 0$

$\left(u + 4\right) = 0 \mathmr{and} \left(u - 3\right) = 0$

$u = - 4 \mathmr{and} u = 3$