Solve 120(r + 4) เมื่อ r r = 10 | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$120 ( r + 4 ) r = 10$$

Solve for r

$r=\frac{7\sqrt{3}}{6}-2\approx 0.020725942$
$r=-\frac{7\sqrt{3}}{6}-2\approx -4.020725942$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Use the distributive property to multiply $120$ by $r+4$.

$$\left(120r+480\right)r=10$$

Use the distributive property to multiply $120r+480$ by $r$.

$$120r^{2}+480r=10$$

Subtract $10$ from both sides.

$$120r^{2}+480r-10=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $120$ for $a$, $480$ for $b$, and $-10$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$r=\frac{-480±\sqrt{480^{2}-4\times 120\left(-10\right)}}{2\times 120}$$

Square $480$.

$$r=\frac{-480±\sqrt{230400-4\times 120\left(-10\right)}}{2\times 120}$$

Multiply $-4$ times $120$.

$$r=\frac{-480±\sqrt{230400-480\left(-10\right)}}{2\times 120}$$

Multiply $-480$ times $-10$.

$$r=\frac{-480±\sqrt{230400+4800}}{2\times 120}$$

Add $230400$ to $4800$.

$$r=\frac{-480±\sqrt{235200}}{2\times 120}$$

Take the square root of $235200$.

$$r=\frac{-480±280\sqrt{3}}{2\times 120}$$

Multiply $2$ times $120$.

$$r=\frac{-480±280\sqrt{3}}{240}$$

Now solve the equation $r=\frac{-480±280\sqrt{3}}{240}$ when $±$ is plus. Add $-480$ to $280\sqrt{3}$.

$$r=\frac{280\sqrt{3}-480}{240}$$

Divide $-480+280\sqrt{3}$ by $240$.

$$r=\frac{7\sqrt{3}}{6}-2$$

Now solve the equation $r=\frac{-480±280\sqrt{3}}{240}$ when $±$ is minus. Subtract $280\sqrt{3}$ from $-480$.

$$r=\frac{-280\sqrt{3}-480}{240}$$

Divide $-480-280\sqrt{3}$ by $240$.

$$r=-\frac{7\sqrt{3}}{6}-2$$

The equation is now solved.

$$r=\frac{7\sqrt{3}}{6}-2$$ $$r=-\frac{7\sqrt{3}}{6}-2$$

Steps for Completing the Square

Use the distributive property to multiply $120$ by $r+4$.

$$\left(120r+480\right)r=10$$

Use the distributive property to multiply $120r+480$ by $r$.

$$120r^{2}+480r=10$$

Divide both sides by $120$.

$$\frac{120r^{2}+480r}{120}=\frac{10}{120}$$

Dividing by $120$ undoes the multiplication by $120$.

$$r^{2}+\frac{480}{120}r=\frac{10}{120}$$

Divide $480$ by $120$.

$$r^{2}+4r=\frac{10}{120}$$

Reduce the fraction $\frac{10}{120}$ to lowest terms by extracting and canceling out $10$.

$$r^{2}+4r=\frac{1}{12}$$

Divide $4$, the coefficient of the $x$ term, by $2$ to get $2$. Then add the square of $2$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$r^{2}+4r+2^{2}=\frac{1}{12}+2^{2}$$

Square $2$.

$$r^{2}+4r+4=\frac{1}{12}+4$$

Add $\frac{1}{12}$ to $4$.

$$r^{2}+4r+4=\frac{49}{12}$$

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

$$\left(r+2\right)^{2}=\frac{49}{12}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(r+2\right)^{2}}=\sqrt{\frac{49}{12}}$$

Simplify.

$$r+2=\frac{7\sqrt{3}}{6}$$ $$r+2=-\frac{7\sqrt{3}}{6}$$

Subtract $2$ from both sides of the equation.

$$r=\frac{7\sqrt{3}}{6}-2$$ $$r=-\frac{7\sqrt{3}}{6}-2$$