How to solve 3(z)^(2)=198+15z

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve 3(z)^(2)=198+15z | Equatix

Question

$$3 { z }^{ 2 } =198+15z$$

Solve for z

$z=-6$
$z=11$

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring By Grouping

Subtract $198$ from both sides.

$$3z^{2}-198=15z$$

Subtract $15z$ from both sides.

$$3z^{2}-198-15z=0$$

Divide both sides by $3$.

$$z^{2}-66-5z=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$z^{2}-5z-66=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $z^{2}+az+bz-66$. To find $a$ and $b$, set up a system to be solved.

$$a+b=-5$$ $$ab=1\left(-66\right)=-66$$

Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product $-66$.

$$1,-66$$ $$2,-33$$ $$3,-22$$ $$6,-11$$

Calculate the sum for each pair.

$$1-66=-65$$ $$2-33=-31$$ $$3-22=-19$$ $$6-11=-5$$

The solution is the pair that gives sum $-5$.

$$a=-11$$ $$b=6$$

Rewrite $z^{2}-5z-66$ as $\left(z^{2}-11z\right)+\left(6z-66\right)$.

$$\left(z^{2}-11z\right)+\left(6z-66\right)$$

Factor out $z$ in the first and $6$ in the second group.

$$z\left(z-11\right)+6\left(z-11\right)$$

Factor out common term $z-11$ by using distributive property.

$$\left(z-11\right)\left(z+6\right)$$

To find equation solutions, solve $z-11=0$ and $z+6=0$.

$$z=11$$ $$z=-6$$

Steps Using the Quadratic Formula

Subtract $198$ from both sides.

$$3z^{2}-198=15z$$

Subtract $15z$ from both sides.

$$3z^{2}-198-15z=0$$

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.

$$3z^{2}-15z-198=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $3$ for $a$, $-15$ for $b$, and $-198$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$z=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 3\left(-198\right)}}{2\times 3}$$

Square $-15$.

$$z=\frac{-\left(-15\right)±\sqrt{225-4\times 3\left(-198\right)}}{2\times 3}$$

Multiply $-4$ times $3$.

$$z=\frac{-\left(-15\right)±\sqrt{225-12\left(-198\right)}}{2\times 3}$$

Multiply $-12$ times $-198$.

$$z=\frac{-\left(-15\right)±\sqrt{225+2376}}{2\times 3}$$

Add $225$ to $2376$.

$$z=\frac{-\left(-15\right)±\sqrt{2601}}{2\times 3}$$

Take the square root of $2601$.

$$z=\frac{-\left(-15\right)±51}{2\times 3}$$

The opposite of $-15$ is $15$.

$$z=\frac{15±51}{2\times 3}$$

Multiply $2$ times $3$.

$$z=\frac{15±51}{6}$$

Now solve the equation $z=\frac{15±51}{6}$ when $±$ is plus. Add $15$ to $51$.

$$z=\frac{66}{6}$$

Divide $66$ by $6$.

$$z=11$$

Now solve the equation $z=\frac{15±51}{6}$ when $±$ is minus. Subtract $51$ from $15$.

$$z=-\frac{36}{6}$$

Divide $-36$ by $6$.

$$z=-6$$

The equation is now solved.

$$z=11$$ $$z=-6$$

Steps for Completing the Square

Subtract $15z$ from both sides.

$$3z^{2}-15z=198$$

Divide both sides by $3$.

$$\frac{3z^{2}-15z}{3}=\frac{198}{3}$$

Dividing by $3$ undoes the multiplication by $3$.

$$z^{2}+\left(-\frac{15}{3}\right)z=\frac{198}{3}$$

Divide $-15$ by $3$.

$$z^{2}-5z=\frac{198}{3}$$

Divide $198$ by $3$.

$$z^{2}-5z=66$$

Divide $-5$, the coefficient of the $x$ term, by $2$ to get $-\frac{5}{2}$. Then add the square of $-\frac{5}{2}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$z^{2}-5z+\left(-\frac{5}{2}\right)^{2}=66+\left(-\frac{5}{2}\right)^{2}$$

Square $-\frac{5}{2}$ by squaring both the numerator and the denominator of the fraction.

$$z^{2}-5z+\frac{25}{4}=66+\frac{25}{4}$$

Add $66$ to $\frac{25}{4}$.

$$z^{2}-5z+\frac{25}{4}=\frac{289}{4}$$

Factor $z^{2}-5z+\frac{25}{4}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(z-\frac{5}{2}\right)^{2}=\frac{289}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(z-\frac{5}{2}\right)^{2}}=\sqrt{\frac{289}{4}}$$

Simplify.

$$z-\frac{5}{2}=\frac{17}{2}$$ $$z-\frac{5}{2}=-\frac{17}{2}$$

Add $\frac{5}{2}$ to both sides of the equation.

$$z=11$$ $$z=-6$$

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Example

Question

Solve for z

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square