Solve x = - 3x ^ 2 + 2x + 7 | 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

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

Solve for x

$x = \frac{\sqrt{85} + 1}{6} \approx 1.70325741$
$x=\frac{1-\sqrt{85}}{6}\approx -1.369924076$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

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

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

Subtract $2x$ from both sides.

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

Combine $x$ and $-2x$ to get $-x$.

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

Subtract $7$ from both sides.

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

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.

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

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $3$ for $a$, $-1$ for $b$, and $-7$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 3\left(-7\right)}}{2\times 3}$$

Multiply $-4$ times $3$.

$$x=\frac{-\left(-1\right)±\sqrt{1-12\left(-7\right)}}{2\times 3}$$

Multiply $-12$ times $-7$.

$$x=\frac{-\left(-1\right)±\sqrt{1+84}}{2\times 3}$$

Add $1$ to $84$.

$$x=\frac{-\left(-1\right)±\sqrt{85}}{2\times 3}$$

The opposite of $-1$ is $1$.

$$x=\frac{1±\sqrt{85}}{2\times 3}$$

Multiply $2$ times $3$.

$$x=\frac{1±\sqrt{85}}{6}$$

Now solve the equation $x=\frac{1±\sqrt{85}}{6}$ when $±$ is plus. Add $1$ to $\sqrt{85}$.

$$x=\frac{\sqrt{85}+1}{6}$$

Now solve the equation $x=\frac{1±\sqrt{85}}{6}$ when $±$ is minus. Subtract $\sqrt{85}$ from $1$.

$$x=\frac{1-\sqrt{85}}{6}$$

The equation is now solved.

$$x=\frac{\sqrt{85}+1}{6}$$ $$x=\frac{1-\sqrt{85}}{6}$$

Steps for Completing the Square

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

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

Subtract $2x$ from both sides.

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

Combine $x$ and $-2x$ to get $-x$.

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

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 $x^{2}+bx=c$.

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

Divide both sides by $3$.

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

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

$$x^{2}-\frac{1}{3}x=\frac{7}{3}$$

Divide $-\frac{1}{3}$, the coefficient of the $x$ term, by $2$ to get $-\frac{1}{6}$. Then add the square of $-\frac{1}{6}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{3}x+\left(-\frac{1}{6}\right)^{2}=\frac{7}{3}+\left(-\frac{1}{6}\right)^{2}$$

Square $-\frac{1}{6}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{7}{3}+\frac{1}{36}$$

Add $\frac{7}{3}$ to $\frac{1}{36}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{85}{36}$$

Factor $x^{2}-\frac{1}{3}x+\frac{1}{36}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{1}{6}\right)^{2}=\frac{85}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{6}\right)^{2}}=\sqrt{\frac{85}{36}}$$

Simplify.

$$x-\frac{1}{6}=\frac{\sqrt{85}}{6}$$ $$x-\frac{1}{6}=-\frac{\sqrt{85}}{6}$$

Add $\frac{1}{6}$ to both sides of the equation.

$$x=\frac{\sqrt{85}+1}{6}$$ $$x=\frac{1-\sqrt{85}}{6}$$