Solve 2(x)^(2)+4x-7x=7(x)^(2) | 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

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

Solve for x

$x=-\frac{3}{5}=-0.6$
$x=0$

Steps Using Factoring

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring

Combine $4x$ and $-7x$ to get $-3x$.

$$2x^{2}-3x=7x^{2}$$

Subtract $7x^{2}$ from both sides.

$$2x^{2}-3x-7x^{2}=0$$

Combine $2x^{2}$ and $-7x^{2}$ to get $-5x^{2}$.

$$-5x^{2}-3x=0$$

Factor out $x$.

$$x\left(-5x-3\right)=0$$

To find equation solutions, solve $x=0$ and $-5x-3=0$.

$$x=0$$ $$x=-\frac{3}{5}$$

Steps Using the Quadratic Formula

Combine $4x$ and $-7x$ to get $-3x$.

$$2x^{2}-3x=7x^{2}$$

Subtract $7x^{2}$ from both sides.

$$2x^{2}-3x-7x^{2}=0$$

Combine $2x^{2}$ and $-7x^{2}$ to get $-5x^{2}$.

$$-5x^{2}-3x=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-5$ for $a$, $-3$ for $b$, and $0$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}}}{2\left(-5\right)}$$

Take the square root of $\left(-3\right)^{2}$.

$$x=\frac{-\left(-3\right)±3}{2\left(-5\right)}$$

The opposite of $-3$ is $3$.

$$x=\frac{3±3}{2\left(-5\right)}$$

Multiply $2$ times $-5$.

$$x=\frac{3±3}{-10}$$

Now solve the equation $x=\frac{3±3}{-10}$ when $±$ is plus. Add $3$ to $3$.

$$x=\frac{6}{-10}$$

Reduce the fraction $\frac{6}{-10}$ to lowest terms by extracting and canceling out $2$.

$$x=-\frac{3}{5}$$

Now solve the equation $x=\frac{3±3}{-10}$ when $±$ is minus. Subtract $3$ from $3$.

$$x=\frac{0}{-10}$$

Divide $0$ by $-10$.

$$x=0$$

The equation is now solved.

$$x=-\frac{3}{5}$$ $$x=0$$

Steps for Completing the Square

Combine $4x$ and $-7x$ to get $-3x$.

$$2x^{2}-3x=7x^{2}$$

Subtract $7x^{2}$ from both sides.

$$2x^{2}-3x-7x^{2}=0$$

Combine $2x^{2}$ and $-7x^{2}$ to get $-5x^{2}$.

$$-5x^{2}-3x=0$$

Divide both sides by $-5$.

$$\frac{-5x^{2}-3x}{-5}=\frac{0}{-5}$$

Dividing by $-5$ undoes the multiplication by $-5$.

$$x^{2}+\left(-\frac{3}{-5}\right)x=\frac{0}{-5}$$

Divide $-3$ by $-5$.

$$x^{2}+\frac{3}{5}x=\frac{0}{-5}$$

Divide $0$ by $-5$.

$$x^{2}+\frac{3}{5}x=0$$

Divide $\frac{3}{5}$, the coefficient of the $x$ term, by $2$ to get $\frac{3}{10}$. Then add the square of $\frac{3}{10}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{3}{5}x+\left(\frac{3}{10}\right)^{2}=\left(\frac{3}{10}\right)^{2}$$

Square $\frac{3}{10}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{5}x+\frac{9}{100}=\frac{9}{100}$$

Factor $x^{2}+\frac{3}{5}x+\frac{9}{100}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+\frac{3}{10}\right)^{2}=\frac{9}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{10}\right)^{2}}=\sqrt{\frac{9}{100}}$$

Simplify.

$$x+\frac{3}{10}=\frac{3}{10}$$ $$x+\frac{3}{10}=-\frac{3}{10}$$

Subtract $\frac{3}{10}$ from both sides of the equation.

$$x=0$$ $$x=-\frac{3}{5}$$