How to solve 5 - 2x ^ 2 = 6x

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 5 - 2x ^ 2 = 6x | Equatix

Question

$$5-2x^{2}=6x$$

Solve for x

$x=\frac{\sqrt{19}-3}{2}\approx 0.679449472$
$x=\frac{-\sqrt{19}-3}{2}\approx -3.679449472$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Subtract $6x$ from both sides.

$$5-2x^{2}-6x=0$$

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.

$$-2x^{2}-6x+5=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-2$ for $a$, $-6$ for $b$, and $5$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-2\right)\times 5}}{2\left(-2\right)}$$

Square $-6$.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\left(-2\right)\times 5}}{2\left(-2\right)}$$

Multiply $-4$ times $-2$.

$$x=\frac{-\left(-6\right)±\sqrt{36+8\times 5}}{2\left(-2\right)}$$

Multiply $8$ times $5$.

$$x=\frac{-\left(-6\right)±\sqrt{36+40}}{2\left(-2\right)}$$

Add $36$ to $40$.

$$x=\frac{-\left(-6\right)±\sqrt{76}}{2\left(-2\right)}$$

Take the square root of $76$.

$$x=\frac{-\left(-6\right)±2\sqrt{19}}{2\left(-2\right)}$$

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

$$x=\frac{6±2\sqrt{19}}{2\left(-2\right)}$$

Multiply $2$ times $-2$.

$$x=\frac{6±2\sqrt{19}}{-4}$$

Now solve the equation $x=\frac{6±2\sqrt{19}}{-4}$ when $±$ is plus. Add $6$ to $2\sqrt{19}$.

$$x=\frac{2\sqrt{19}+6}{-4}$$

Divide $6+2\sqrt{19}$ by $-4$.

$$x=\frac{-\sqrt{19}-3}{2}$$

Now solve the equation $x=\frac{6±2\sqrt{19}}{-4}$ when $±$ is minus. Subtract $2\sqrt{19}$ from $6$.

$$x=\frac{6-2\sqrt{19}}{-4}$$

Divide $6-2\sqrt{19}$ by $-4$.

$$x=\frac{\sqrt{19}-3}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{19}-3}{2}$$ $$x=\frac{\sqrt{19}-3}{2}$$

Steps for Completing the Square

Subtract $6x$ from both sides.

$$5-2x^{2}-6x=0$$

Subtract $5$ from both sides. Anything subtracted from zero gives its negation.

$$-2x^{2}-6x=-5$$

Divide both sides by $-2$.

$$\frac{-2x^{2}-6x}{-2}=-\frac{5}{-2}$$

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

$$x^{2}+\left(-\frac{6}{-2}\right)x=-\frac{5}{-2}$$

Divide $-6$ by $-2$.

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

Divide $-5$ by $-2$.

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

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

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

Square $\frac{3}{2}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=\frac{5}{2}+\frac{9}{4}$$

Add $\frac{5}{2}$ to $\frac{9}{4}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+3x+\frac{9}{4}=\frac{19}{4}$$

Factor $x^{2}+3x+\frac{9}{4}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{19}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{19}{4}}$$

Simplify.

$$x+\frac{3}{2}=\frac{\sqrt{19}}{2}$$ $$x+\frac{3}{2}=-\frac{\sqrt{19}}{2}$$

Subtract $\frac{3}{2}$ from both sides of the equation.

$$x=\frac{\sqrt{19}-3}{2}$$ $$x=\frac{-\sqrt{19}-3}{2}$$

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Example

Question

Solve for x

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Steps for Completing the Square