How to solve 5x(3x-2)=2(x+3)-3

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 5x(3x-2)=2(x+3)-3 | Equatix

Question

$$5x(3x-2)=2(x+3)-3$$

Solve for x

$x=-\frac{1}{5}=-0.2$
$x=1$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Use the distributive property to multiply $5x$ by $3x-2$.

$$15x^{2}-10x=2\left(x+3\right)-3$$

Use the distributive property to multiply $2$ by $x+3$.

$$15x^{2}-10x=2x+6-3$$

Subtract $3$ from $6$ to get $3$.

$$15x^{2}-10x=2x+3$$

Subtract $2x$ from both sides.

$$15x^{2}-10x-2x=3$$

Combine $-10x$ and $-2x$ to get $-12x$.

$$15x^{2}-12x=3$$

Subtract $3$ from both sides.

$$15x^{2}-12x-3=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $15$ for $a$, $-12$ for $b$, and $-3$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 15\left(-3\right)}}{2\times 15}$$

Square $-12$.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 15\left(-3\right)}}{2\times 15}$$

Multiply $-4$ times $15$.

$$x=\frac{-\left(-12\right)±\sqrt{144-60\left(-3\right)}}{2\times 15}$$

Multiply $-60$ times $-3$.

$$x=\frac{-\left(-12\right)±\sqrt{144+180}}{2\times 15}$$

Add $144$ to $180$.

$$x=\frac{-\left(-12\right)±\sqrt{324}}{2\times 15}$$

Take the square root of $324$.

$$x=\frac{-\left(-12\right)±18}{2\times 15}$$

The opposite of $-12$ is $12$.

$$x=\frac{12±18}{2\times 15}$$

Multiply $2$ times $15$.

$$x=\frac{12±18}{30}$$

Now solve the equation $x=\frac{12±18}{30}$ when $±$ is plus. Add $12$ to $18$.

$$x=\frac{30}{30}$$

Divide $30$ by $30$.

$$x=1$$

Now solve the equation $x=\frac{12±18}{30}$ when $±$ is minus. Subtract $18$ from $12$.

$$x=-\frac{6}{30}$$

Reduce the fraction $\frac{-6}{30}$ to lowest terms by extracting and canceling out $6$.

$$x=-\frac{1}{5}$$

The equation is now solved.

$$x=1$$ $$x=-\frac{1}{5}$$

Steps for Completing the Square

Use the distributive property to multiply $5x$ by $3x-2$.

$$15x^{2}-10x=2\left(x+3\right)-3$$

Use the distributive property to multiply $2$ by $x+3$.

$$15x^{2}-10x=2x+6-3$$

Subtract $3$ from $6$ to get $3$.

$$15x^{2}-10x=2x+3$$

Subtract $2x$ from both sides.

$$15x^{2}-10x-2x=3$$

Combine $-10x$ and $-2x$ to get $-12x$.

$$15x^{2}-12x=3$$

Divide both sides by $15$.

$$\frac{15x^{2}-12x}{15}=\frac{3}{15}$$

Dividing by $15$ undoes the multiplication by $15$.

$$x^{2}+\left(-\frac{12}{15}\right)x=\frac{3}{15}$$

Reduce the fraction $\frac{-12}{15}$ to lowest terms by extracting and canceling out $3$.

$$x^{2}-\frac{4}{5}x=\frac{3}{15}$$

Reduce the fraction $\frac{3}{15}$ to lowest terms by extracting and canceling out $3$.

$$x^{2}-\frac{4}{5}x=\frac{1}{5}$$

Divide $-\frac{4}{5}$, the coefficient of the $x$ term, by $2$ to get $-\frac{2}{5}$. Then add the square of $-\frac{2}{5}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{4}{5}x+\left(-\frac{2}{5}\right)^{2}=\frac{1}{5}+\left(-\frac{2}{5}\right)^{2}$$

Square $-\frac{2}{5}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{4}{5}x+\frac{4}{25}=\frac{1}{5}+\frac{4}{25}$$

Add $\frac{1}{5}$ to $\frac{4}{25}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{4}{5}x+\frac{4}{25}=\frac{9}{25}$$

Factor $x^{2}-\frac{4}{5}x+\frac{4}{25}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{2}{5}\right)^{2}=\frac{9}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{5}\right)^{2}}=\sqrt{\frac{9}{25}}$$

Simplify.

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

Add $\frac{2}{5}$ to both sides of the equation.

$$x=1$$ $$x=-\frac{1}{5}$$

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Solve for x

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Steps for Completing the Square