Solve for z square root of 280+40z-z=17

$\sqrt{280 + 40 z} - z = 17$

Add $z$ to both sides of the equation.

$\sqrt{280 + 40 z} = 17 + z$

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

${\sqrt{280 + 40 z}}^{2} = {(17 + z)}^{2}$

Simplify each side of the equation.

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Multiply the exponents in ${({(280 + 40 z)}^{\frac{1}{2}})}^{2}$ .

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

${(280 + 40 z)}^{\frac{1}{2} \cdot 2} = {(17 + z)}^{2}$

Cancel the common factor of $2$ .

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Cancel the common factor.

${(280 + 40 z)}^{\frac{1}{2} \cdot 2} = {(17 + z)}^{2}$

Rewrite the expression.

${(280 + 40 z)}^{1} = {(17 + z)}^{2}$

Simplify.

$280 + 40 z = {(17 + z)}^{2}$

Solve for $z$ .

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Simplify ${(17 + z)}^{2}$ .

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Rewrite ${(17 + z)}^{2}$ as $(17 + z) (17 + z)$ .

$280 + 40 z = (17 + z) (17 + z)$

Expand $(17 + z) (17 + z)$ using the FOIL Method.

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Apply the distributive property.

$280 + 40 z = 17 (17 + z) + z (17 + z)$

Apply the distributive property.

$280 + 40 z = 17 \cdot 17 + 17 z + z (17 + z)$

Apply the distributive property.

$280 + 40 z = 17 \cdot 17 + 17 z + z \cdot 17 + z \cdot z$

Simplify and combine like terms.

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Simplify each term.

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Multiply $17$ by $17$ .

$280 + 40 z = 289 + 17 z + z \cdot 17 + z \cdot z$

Move $17$ to the left of $z$ .

$280 + 40 z = 289 + 17 z + 17 \cdot z + z \cdot z$

Multiply $z$ by $z$ .

$280 + 40 z = 289 + 17 z + 17 z + z^{2}$

Add $17 z$ and $17 z$ .

$280 + 40 z = 289 + 34 z + z^{2}$

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

$289 + 34 z + z^{2} = 280 + 40 z$

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

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Subtract $40 z$ from both sides of the equation.

$289 + 34 z + z^{2} - 40 z = 280$

Subtract $40 z$ from $34 z$ .

$289 + z^{2} - 6 z = 280$

Move $280$ to the left side of the equation by subtracting it from both sides.

$289 + z^{2} - 6 z - 280 = 0$

Subtract $280$ from $289$ .

$z^{2} - 6 z + 9 = 0$

Factor using the perfect square rule.

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Rewrite $9$ as $3^{2}$ .

$z^{2} - 6 z + 3^{2} = 0$

Check the middle term by multiplying $2 a b$ and compare this result with the middle term in the original expression.

$2 a b = 2 \cdot z \cdot - 3$

Simplify.

$2 a b = - 6 z$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = z$ and $b = - 3$ .

${(z - 3)}^{2} = 0$

Set the $z - 3$ equal to $0$ .

$z - 3 = 0$

Add $3$ to both sides of the equation.

$z = 3$

Solve for z square root of 280+40z-z=17

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