Solve Solve 5x + 6 = 3(10 - x) for x. | 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

$$5 x + 6 = 3 ( 10 - x ) x$$

Solve for x

$x = \frac{\sqrt{553} + 25}{6} \approx 8.085992005$
$x=\frac{25-\sqrt{553}}{6}\approx 0.247341328$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Use the distributive property to multiply $3$ by $10-x$.

$$5x+6=\left(30-3x\right)x$$

Use the distributive property to multiply $30-3x$ by $x$.

$$5x+6=30x-3x^{2}$$

Subtract $30x$ from both sides.

$$5x+6-30x=-3x^{2}$$

Combine $5x$ and $-30x$ to get $-25x$.

$$-25x+6=-3x^{2}$$

Add $3x^{2}$ to both sides.

$$-25x+6+3x^{2}=0$$

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.

$$3x^{2}-25x+6=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $3$ for $a$, $-25$ for $b$, and $6$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-25\right)±\sqrt{\left(-25\right)^{2}-4\times 3\times 6}}{2\times 3}$$

Square $-25$.

$$x=\frac{-\left(-25\right)±\sqrt{625-4\times 3\times 6}}{2\times 3}$$

Multiply $-4$ times $3$.

$$x=\frac{-\left(-25\right)±\sqrt{625-12\times 6}}{2\times 3}$$

Multiply $-12$ times $6$.

$$x=\frac{-\left(-25\right)±\sqrt{625-72}}{2\times 3}$$

Add $625$ to $-72$.

$$x=\frac{-\left(-25\right)±\sqrt{553}}{2\times 3}$$

The opposite of $-25$ is $25$.

$$x=\frac{25±\sqrt{553}}{2\times 3}$$

Multiply $2$ times $3$.

$$x=\frac{25±\sqrt{553}}{6}$$

Now solve the equation $x=\frac{25±\sqrt{553}}{6}$ when $±$ is plus. Add $25$ to $\sqrt{553}$.

$$x=\frac{\sqrt{553}+25}{6}$$

Now solve the equation $x=\frac{25±\sqrt{553}}{6}$ when $±$ is minus. Subtract $\sqrt{553}$ from $25$.

$$x=\frac{25-\sqrt{553}}{6}$$

The equation is now solved.

$$x=\frac{\sqrt{553}+25}{6}$$ $$x=\frac{25-\sqrt{553}}{6}$$

Steps for Completing the Square

Use the distributive property to multiply $3$ by $10-x$.

$$5x+6=\left(30-3x\right)x$$

Use the distributive property to multiply $30-3x$ by $x$.

$$5x+6=30x-3x^{2}$$

Subtract $30x$ from both sides.

$$5x+6-30x=-3x^{2}$$

Combine $5x$ and $-30x$ to get $-25x$.

$$-25x+6=-3x^{2}$$

Add $3x^{2}$ to both sides.

$$-25x+6+3x^{2}=0$$

Subtract $6$ from both sides. Anything subtracted from zero gives its negation.

$$-25x+3x^{2}=-6$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form $x^{2}+bx=c$.

$$3x^{2}-25x=-6$$

Divide both sides by $3$.

$$\frac{3x^{2}-25x}{3}=-\frac{6}{3}$$

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

$$x^{2}-\frac{25}{3}x=-\frac{6}{3}$$

Divide $-6$ by $3$.

$$x^{2}-\frac{25}{3}x=-2$$

Divide $-\frac{25}{3}$, the coefficient of the $x$ term, by $2$ to get $-\frac{25}{6}$. Then add the square of $-\frac{25}{6}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{25}{3}x+\left(-\frac{25}{6}\right)^{2}=-2+\left(-\frac{25}{6}\right)^{2}$$

Square $-\frac{25}{6}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{25}{3}x+\frac{625}{36}=-2+\frac{625}{36}$$

Add $-2$ to $\frac{625}{36}$.

$$x^{2}-\frac{25}{3}x+\frac{625}{36}=\frac{553}{36}$$

Factor $x^{2}-\frac{25}{3}x+\frac{625}{36}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{25}{6}\right)^{2}=\frac{553}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{25}{6}\right)^{2}}=\sqrt{\frac{553}{36}}$$

Simplify.

$$x-\frac{25}{6}=\frac{\sqrt{553}}{6}$$ $$x-\frac{25}{6}=-\frac{\sqrt{553}}{6}$$

Add $\frac{25}{6}$ to both sides of the equation.

$$x=\frac{\sqrt{553}+25}{6}$$ $$x=\frac{25-\sqrt{553}}{6}$$