Solve 2x(x - 3) + 5(x - 3) = 0 | 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

$$2x(x-3)+5(x-3)=0$$

Solve for x

$x = -\frac{5}{2} = -2\frac{1}{2} = -2.5$
$x=3$

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring By Grouping

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

$$2x^{2}-6x+5\left(x-3\right)=0$$

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

$$2x^{2}-6x+5x-15=0$$

Combine $-6x$ and $5x$ to get $-x$.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $2x^{2}+ax+bx-15$. To find $a$ and $b$, set up a system to be solved.

$$a+b=-1$$ $$ab=2\left(-15\right)=-30$$

Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product $-30$.

$$1,-30$$ $$2,-15$$ $$3,-10$$ $$5,-6$$

Calculate the sum for each pair.

$$1-30=-29$$ $$2-15=-13$$ $$3-10=-7$$ $$5-6=-1$$

The solution is the pair that gives sum $-1$.

$$a=-6$$ $$b=5$$

Rewrite $2x^{2}-x-15$ as $\left(2x^{2}-6x\right)+\left(5x-15\right)$.

$$\left(2x^{2}-6x\right)+\left(5x-15\right)$$

Factor out $2x$ in the first and $5$ in the second group.

$$2x\left(x-3\right)+5\left(x-3\right)$$

Factor out common term $x-3$ by using distributive property.

$$\left(x-3\right)\left(2x+5\right)$$

To find equation solutions, solve $x-3=0$ and $2x+5=0$.

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

Steps Using the Quadratic Formula

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

$$2x^{2}-6x+5\left(x-3\right)=0$$

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

$$2x^{2}-6x+5x-15=0$$

Combine $-6x$ and $5x$ to get $-x$.

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

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $2$ for $a$, $-1$ for $b$, and $-15$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 2\left(-15\right)}}{2\times 2}$$

Multiply $-4$ times $2$.

$$x=\frac{-\left(-1\right)±\sqrt{1-8\left(-15\right)}}{2\times 2}$$

Multiply $-8$ times $-15$.

$$x=\frac{-\left(-1\right)±\sqrt{1+120}}{2\times 2}$$

Add $1$ to $120$.

$$x=\frac{-\left(-1\right)±\sqrt{121}}{2\times 2}$$

Take the square root of $121$.

$$x=\frac{-\left(-1\right)±11}{2\times 2}$$

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

$$x=\frac{1±11}{2\times 2}$$

Multiply $2$ times $2$.

$$x=\frac{1±11}{4}$$

Now solve the equation $x=\frac{1±11}{4}$ when $±$ is plus. Add $1$ to $11$.

$$x=\frac{12}{4}$$

Divide $12$ by $4$.

$$x=3$$

Now solve the equation $x=\frac{1±11}{4}$ when $±$ is minus. Subtract $11$ from $1$.

$$x=-\frac{10}{4}$$

Reduce the fraction $\frac{-10}{4}$ to lowest terms by extracting and canceling out $2$.

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

The equation is now solved.

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

Steps for Completing the Square

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

$$2x^{2}-6x+5\left(x-3\right)=0$$

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

$$2x^{2}-6x+5x-15=0$$

Combine $-6x$ and $5x$ to get $-x$.

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

Add $15$ to both sides. Anything plus zero gives itself.

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

Divide both sides by $2$.

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

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

$$x^{2}-\frac{1}{2}x=\frac{15}{2}$$

Divide $-\frac{1}{2}$, the coefficient of the $x$ term, by $2$ to get $-\frac{1}{4}$. Then add the square of $-\frac{1}{4}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=\frac{15}{2}+\left(-\frac{1}{4}\right)^{2}$$

Square $-\frac{1}{4}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{15}{2}+\frac{1}{16}$$

Add $\frac{15}{2}$ to $\frac{1}{16}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{121}{16}$$

Factor $x^{2}-\frac{1}{2}x+\frac{1}{16}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{1}{4}\right)^{2}=\frac{121}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{\frac{121}{16}}$$

Simplify.

$$x-\frac{1}{4}=\frac{11}{4}$$ $$x-\frac{1}{4}=-\frac{11}{4}$$

Add $\frac{1}{4}$ to both sides of the equation.

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