Solve for z square root of -6z+3+z=-4

$\sqrt{- 6 z + 3} + z = - 4$

Subtract $z$ from both sides of the equation.

$\sqrt{- 6 z + 3} = - 4 - z$

To remove the radical on the left side of the equation, square both sides of the equation.

${\sqrt{- 6 z + 3}}^{2} = {(- 4 - z)}^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{- 6 z + 3}$ as ${(- 6 z + 3)}^{\frac{1}{2}}$ .

${({(- 6 z + 3)}^{\frac{1}{2}})}^{2} = {(- 4 - z)}^{2}$

Simplify the left side.

Simplify ${({(- 6 z + 3)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(- 6 z + 3)}^{\frac{1}{2}})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${(- 6 z + 3)}^{\frac{1}{2} \cdot 2} = {(- 4 - z)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(- 6 z + 3)}^{\frac{1}{2} \cdot 2} = {(- 4 - z)}^{2}$

Rewrite the expression.

${(- 6 z + 3)}^{1} = {(- 4 - z)}^{2}$

Simplify.

$- 6 z + 3 = {(- 4 - z)}^{2}$

Simplify the right side.

Simplify ${(- 4 - z)}^{2}$ .

Rewrite ${(- 4 - z)}^{2}$ as $(- 4 - z) (- 4 - z)$ .

$- 6 z + 3 = (- 4 - z) (- 4 - z)$

Expand $(- 4 - z) (- 4 - z)$ using the FOIL Method.

Apply the distributive property.

$- 6 z + 3 = - 4 (- 4 - z) - z (- 4 - z)$

Apply the distributive property.

$- 6 z + 3 = - 4 \cdot - 4 - 4 (- z) - z (- 4 - z)$

Apply the distributive property.

$- 6 z + 3 = - 4 \cdot - 4 - 4 (- z) - z \cdot - 4 - z (- z)$

Simplify and combine like terms.

Simplify each term.

Multiply $- 4$ by $- 4$ .

$- 6 z + 3 = 16 - 4 (- z) - z \cdot - 4 - z (- z)$

Multiply $- 1$ by $- 4$ .

$- 6 z + 3 = 16 + 4 z - z \cdot - 4 - z (- z)$

Multiply $- 4$ by $- 1$ .

$- 6 z + 3 = 16 + 4 z + 4 z - z (- z)$

Rewrite using the commutative property of multiplication.

$- 6 z + 3 = 16 + 4 z + 4 z - 1 \cdot - 1 z \cdot z$

Multiply $z$ by $z$ by adding the exponents.

Move $z$ .

$- 6 z + 3 = 16 + 4 z + 4 z - 1 \cdot - 1 (z \cdot z)$

Multiply $z$ by $z$ .

$- 6 z + 3 = 16 + 4 z + 4 z - 1 \cdot - 1 z^{2}$

Multiply $- 1$ by $- 1$ .

$- 6 z + 3 = 16 + 4 z + 4 z + 1 z^{2}$

Multiply $z^{2}$ by $1$ .

$- 6 z + 3 = 16 + 4 z + 4 z + z^{2}$

Add $4 z$ and $4 z$ .

$- 6 z + 3 = 16 + 8 z + z^{2}$

Solve for $z$ .

Since $z$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$16 + 8 z + z^{2} = - 6 z + 3$

Move all terms containing $z$ to the left side of the equation.

Add $6 z$ to both sides of the equation.

$16 + 8 z + z^{2} + 6 z = 3$

Add $8 z$ and $6 z$ .

$16 + z^{2} + 14 z = 3$

Subtract $3$ from both sides of the equation.

$16 + z^{2} + 14 z - 3 = 0$

Subtract $3$ from $16$ .

$z^{2} + 14 z + 13 = 0$

Factor $z^{2} + 14 z + 13$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $13$ and whose sum is $14$ .

$1, 13$

Write the factored form using these integers.

$(z + 1) (z + 13) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$z + 1 = 0$

$z + 13 = 0$

Set $z + 1$ equal to $0$ and solve for $z$ .

Set $z + 1$ equal to $0$ .

$z + 1 = 0$

Subtract $1$ from both sides of the equation.

$z = - 1$

Set $z + 13$ equal to $0$ and solve for $z$ .

Set $z + 13$ equal to $0$ .

$z + 13 = 0$

Subtract $13$ from both sides of the equation.

$z = - 13$

The final solution is all the values that make $(z + 1) (z + 13) = 0$ true.

$z = - 1, - 13$

Exclude the solutions that do not make $\sqrt{- 6 z + 3} + z = - 4$ true.

$z = - 13$

Solve for z square root of -6z+3+z=-4

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