Solve (x-3)(2x+1)=x.(x+5) | 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

$$(x-3)(2x+1)=x.(x+5)$$

Solve for x

$x = \frac{\sqrt{15} + 3}{2} \approx 3.436491673$
$x=\frac{3-\sqrt{15}}{2}\approx -0.436491673$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Use the distributive property to multiply $x-3$ by $2x+1$ and combine like terms.

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

Subtract $x$ from both sides.

$$2x^{2}-5x-3-x=0$$

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

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

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

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

Square $-6$.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 2\left(-3\right)}}{2\times 2}$$

Multiply $-4$ times $2$.

$$x=\frac{-\left(-6\right)±\sqrt{36-8\left(-3\right)}}{2\times 2}$$

Multiply $-8$ times $-3$.

$$x=\frac{-\left(-6\right)±\sqrt{36+24}}{2\times 2}$$

Add $36$ to $24$.

$$x=\frac{-\left(-6\right)±\sqrt{60}}{2\times 2}$$

Take the square root of $60$.

$$x=\frac{-\left(-6\right)±2\sqrt{15}}{2\times 2}$$

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

$$x=\frac{6±2\sqrt{15}}{2\times 2}$$

Multiply $2$ times $2$.

$$x=\frac{6±2\sqrt{15}}{4}$$

Now solve the equation $x=\frac{6±2\sqrt{15}}{4}$ when $±$ is plus. Add $6$ to $2\sqrt{15}$.

$$x=\frac{2\sqrt{15}+6}{4}$$

Divide $6+2\sqrt{15}$ by $4$.

$$x=\frac{\sqrt{15}+3}{2}$$

Now solve the equation $x=\frac{6±2\sqrt{15}}{4}$ when $±$ is minus. Subtract $2\sqrt{15}$ from $6$.

$$x=\frac{6-2\sqrt{15}}{4}$$

Divide $6-2\sqrt{15}$ by $4$.

$$x=\frac{3-\sqrt{15}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{15}+3}{2}$$ $$x=\frac{3-\sqrt{15}}{2}$$

Steps for Completing the Square

Use the distributive property to multiply $x-3$ by $2x+1$ and combine like terms.

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

Subtract $x$ from both sides.

$$2x^{2}-5x-3-x=0$$

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

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

Add $3$ to both sides. Anything plus zero gives itself.

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

Divide both sides by $2$.

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

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

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

Divide $-6$ by $2$.

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

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.

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

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

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

Add $\frac{3}{2}$ to $\frac{9}{4}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-3x+\frac{9}{4}=\frac{15}{4}$$

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

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-\frac{3}{2}=\frac{\sqrt{15}}{2}$$ $$x-\frac{3}{2}=-\frac{\sqrt{15}}{2}$$

Add $\frac{3}{2}$ to both sides of the equation.

$$x=\frac{\sqrt{15}+3}{2}$$ $$x=\frac{3-\sqrt{15}}{2}$$