Solve 7(x)^(2)+18x=10x+12x | 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

$$7 { x }^{ 2 } +18x=10x+12x$$

Solve for x

$x=\frac{4}{7}\approx 0.571428571$
$x=0$

Steps Using Factoring

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring

Combine $10x$ and $12x$ to get $22x$.

$$7x^{2}+18x=22x$$

Subtract $22x$ from both sides.

$$7x^{2}+18x-22x=0$$

Combine $18x$ and $-22x$ to get $-4x$.

$$7x^{2}-4x=0$$

Factor out $x$.

$$x\left(7x-4\right)=0$$

To find equation solutions, solve $x=0$ and $7x-4=0$.

$$x=0$$ $$x=\frac{4}{7}$$

Steps Using the Quadratic Formula

Combine $10x$ and $12x$ to get $22x$.

$$7x^{2}+18x=22x$$

Subtract $22x$ from both sides.

$$7x^{2}+18x-22x=0$$

Combine $18x$ and $-22x$ to get $-4x$.

$$7x^{2}-4x=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $7$ for $a$, $-4$ for $b$, and $0$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}}}{2\times 7}$$

Take the square root of $\left(-4\right)^{2}$.

$$x=\frac{-\left(-4\right)±4}{2\times 7}$$

The opposite of $-4$ is $4$.

$$x=\frac{4±4}{2\times 7}$$

Multiply $2$ times $7$.

$$x=\frac{4±4}{14}$$

Now solve the equation $x=\frac{4±4}{14}$ when $±$ is plus. Add $4$ to $4$.

$$x=\frac{8}{14}$$

Reduce the fraction $\frac{8}{14}$ to lowest terms by extracting and canceling out $2$.

$$x=\frac{4}{7}$$

Now solve the equation $x=\frac{4±4}{14}$ when $±$ is minus. Subtract $4$ from $4$.

$$x=\frac{0}{14}$$

Divide $0$ by $14$.

$$x=0$$

The equation is now solved.

$$x=\frac{4}{7}$$ $$x=0$$

Steps for Completing the Square

Combine $10x$ and $12x$ to get $22x$.

$$7x^{2}+18x=22x$$

Subtract $22x$ from both sides.

$$7x^{2}+18x-22x=0$$

Combine $18x$ and $-22x$ to get $-4x$.

$$7x^{2}-4x=0$$

Divide both sides by $7$.

$$\frac{7x^{2}-4x}{7}=\frac{0}{7}$$

Dividing by $7$ undoes the multiplication by $7$.

$$x^{2}-\frac{4}{7}x=\frac{0}{7}$$

Divide $0$ by $7$.

$$x^{2}-\frac{4}{7}x=0$$

Divide $-\frac{4}{7}$, the coefficient of the $x$ term, by $2$ to get $-\frac{2}{7}$. Then add the square of $-\frac{2}{7}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{4}{7}x+\left(-\frac{2}{7}\right)^{2}=\left(-\frac{2}{7}\right)^{2}$$

Square $-\frac{2}{7}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{4}{7}x+\frac{4}{49}=\frac{4}{49}$$

Factor $x^{2}-\frac{4}{7}x+\frac{4}{49}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{2}{7}\right)^{2}=\frac{4}{49}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{7}\right)^{2}}=\sqrt{\frac{4}{49}}$$

Simplify.

$$x-\frac{2}{7}=\frac{2}{7}$$ $$x-\frac{2}{7}=-\frac{2}{7}$$

Add $\frac{2}{7}$ to both sides of the equation.

$$x=\frac{4}{7}$$ $$x=0$$