How to solve 2x ^ 2 = 6 - 3x

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve 2x ^ 2 = 6 - 3x | Equatix

Question

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

Solve for x

$x = \frac{\sqrt{57} - 3}{4} \approx 1.137458609$
$x=\frac{-\sqrt{57}-3}{4}\approx -2.637458609$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Subtract $6$ from both sides.

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

Add $3x$ to both sides.

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

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.

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

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

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

Square $3$.

$$x=\frac{-3±\sqrt{9-4\times 2\left(-6\right)}}{2\times 2}$$

Multiply $-4$ times $2$.

$$x=\frac{-3±\sqrt{9-8\left(-6\right)}}{2\times 2}$$

Multiply $-8$ times $-6$.

$$x=\frac{-3±\sqrt{9+48}}{2\times 2}$$

Add $9$ to $48$.

$$x=\frac{-3±\sqrt{57}}{2\times 2}$$

Multiply $2$ times $2$.

$$x=\frac{-3±\sqrt{57}}{4}$$

Now solve the equation $x=\frac{-3±\sqrt{57}}{4}$ when $±$ is plus. Add $-3$ to $\sqrt{57}$.

$$x=\frac{\sqrt{57}-3}{4}$$

Now solve the equation $x=\frac{-3±\sqrt{57}}{4}$ when $±$ is minus. Subtract $\sqrt{57}$ from $-3$.

$$x=\frac{-\sqrt{57}-3}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{57}-3}{4}$$ $$x=\frac{-\sqrt{57}-3}{4}$$

Steps for Completing the Square

Add $3x$ to both sides.

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

Divide both sides by $2$.

$$\frac{2x^{2}+3x}{2}=\frac{6}{2}$$

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

$$x^{2}+\frac{3}{2}x=\frac{6}{2}$$

Divide $6$ by $2$.

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

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.

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

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

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

Add $3$ to $\frac{9}{16}$.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{57}{16}$$

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 $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+\frac{3}{4}\right)^{2}=\frac{57}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{\frac{57}{16}}$$

Simplify.

$$x+\frac{3}{4}=\frac{\sqrt{57}}{4}$$ $$x+\frac{3}{4}=-\frac{\sqrt{57}}{4}$$

Subtract $\frac{3}{4}$ from both sides of the equation.

$$x=\frac{\sqrt{57}-3}{4}$$ $$x=\frac{-\sqrt{57}-3}{4}$$

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Question

Solve for x

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Steps for Completing the Square