Solve 2(x)^(2)=-3x-4--6x-6 | 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

$$2 { x }^{ 2 } =-3x-4--6x-6$$

Solve for x (complex solution)

$x=\frac{3+\sqrt{71}i}{4}\approx 0.75+2.106537443i$
$x=\frac{-\sqrt{71}i+3}{4}\approx 0.75-2.106537443i$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

The opposite of $-6x$ is $6x$.

$$2x^{2}=-3x-4+6x-6$$

Combine $-3x$ and $6x$ to get $3x$.

$$2x^{2}=3x-4-6$$

Subtract $6$ from $-4$ to get $-10$.

$$2x^{2}=3x-10$$

Subtract $3x$ from both sides.

$$2x^{2}-3x=-10$$

Add $10$ to both sides.

$$2x^{2}-3x+10=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $2$ for $a$, $-3$ for $b$, and $10$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 2\times 10}}{2\times 2}$$

Square $-3$.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 2\times 10}}{2\times 2}$$

Multiply $-4$ times $2$.

$$x=\frac{-\left(-3\right)±\sqrt{9-8\times 10}}{2\times 2}$$

Multiply $-8$ times $10$.

$$x=\frac{-\left(-3\right)±\sqrt{9-80}}{2\times 2}$$

Add $9$ to $-80$.

$$x=\frac{-\left(-3\right)±\sqrt{-71}}{2\times 2}$$

Take the square root of $-71$.

$$x=\frac{-\left(-3\right)±\sqrt{71}i}{2\times 2}$$

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

$$x=\frac{3±\sqrt{71}i}{2\times 2}$$

Multiply $2$ times $2$.

$$x=\frac{3±\sqrt{71}i}{4}$$

Now solve the equation $x=\frac{3±\sqrt{71}i}{4}$ when $±$ is plus. Add $3$ to $i\sqrt{71}$.

$$x=\frac{3+\sqrt{71}i}{4}$$

Now solve the equation $x=\frac{3±\sqrt{71}i}{4}$ when $±$ is minus. Subtract $i\sqrt{71}$ from $3$.

$$x=\frac{-\sqrt{71}i+3}{4}$$

The equation is now solved.

$$x=\frac{3+\sqrt{71}i}{4}$$ $$x=\frac{-\sqrt{71}i+3}{4}$$

Steps for Completing the Square

The opposite of $-6x$ is $6x$.

$$2x^{2}=-3x-4+6x-6$$

Combine $-3x$ and $6x$ to get $3x$.

$$2x^{2}=3x-4-6$$

Subtract $6$ from $-4$ to get $-10$.

$$2x^{2}=3x-10$$

Subtract $3x$ from both sides.

$$2x^{2}-3x=-10$$

Divide both sides by $2$.

$$\frac{2x^{2}-3x}{2}=-\frac{10}{2}$$

Dividing by $2$ undoes the multiplication by $2$.

$$x^{2}-\frac{3}{2}x=-\frac{10}{2}$$

Divide $-10$ by $2$.

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

Divide $-\frac{3}{2}$, the coefficient of the $x$ term, by $2$ to get $-\frac{3}{4}$. Then add the square of $-\frac{3}{4}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{2}x+\left(-\frac{3}{4}\right)^{2}=-5+\left(-\frac{3}{4}\right)^{2}$$

Square $-\frac{3}{4}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=-5+\frac{9}{16}$$

Add $-5$ to $\frac{9}{16}$.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=-\frac{71}{16}$$

Factor $x^{2}-\frac{3}{2}x+\frac{9}{16}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{3}{4}\right)^{2}=-\frac{71}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{-\frac{71}{16}}$$

Simplify.

$$x-\frac{3}{4}=\frac{\sqrt{71}i}{4}$$ $$x-\frac{3}{4}=-\frac{\sqrt{71}i}{4}$$

Add $\frac{3}{4}$ to both sides of the equation.

$$x=\frac{3+\sqrt{71}i}{4}$$ $$x=\frac{-\sqrt{71}i+3}{4}$$