Solve 2x(x + 4) = (x - 3)(x - 3) | 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

$$2x(x+4)=(x-3)(x-3)$$

Solve for x (complex solution)

$x=\sqrt{58}-7\approx 0.615773106$
$x=-\left(\sqrt{58}+7\right)\approx -14.615773106$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Multiply $x-3$ and $x-3$ to get $\left(x-3\right)^{2}$.

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

Use the distributive property to multiply $2x$ by $x+4$.

$$2x^{2}+8x=\left(x-3\right)^{2}$$

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-3\right)^{2}$.

$$2x^{2}+8x=x^{2}-6x+9$$

Subtract $x^{2}$ from both sides.

$$2x^{2}+8x-x^{2}=-6x+9$$

Combine $2x^{2}$ and $-x^{2}$ to get $x^{2}$.

$$x^{2}+8x=-6x+9$$

Add $6x$ to both sides.

$$x^{2}+8x+6x=9$$

Combine $8x$ and $6x$ to get $14x$.

$$x^{2}+14x=9$$

Subtract $9$ from both sides.

$$x^{2}+14x-9=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $14$ for $b$, and $-9$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-14±\sqrt{14^{2}-4\left(-9\right)}}{2}$$

Square $14$.

$$x=\frac{-14±\sqrt{196-4\left(-9\right)}}{2}$$

Multiply $-4$ times $-9$.

$$x=\frac{-14±\sqrt{196+36}}{2}$$

Add $196$ to $36$.

$$x=\frac{-14±\sqrt{232}}{2}$$

Take the square root of $232$.

$$x=\frac{-14±2\sqrt{58}}{2}$$

Now solve the equation $x=\frac{-14±2\sqrt{58}}{2}$ when $±$ is plus. Add $-14$ to $2\sqrt{58}$.

$$x=\frac{2\sqrt{58}-14}{2}$$

Divide $-14+2\sqrt{58}$ by $2$.

$$x=\sqrt{58}-7$$

Now solve the equation $x=\frac{-14±2\sqrt{58}}{2}$ when $±$ is minus. Subtract $2\sqrt{58}$ from $-14$.

$$x=\frac{-2\sqrt{58}-14}{2}$$

Divide $-14-2\sqrt{58}$ by $2$.

$$x=-\sqrt{58}-7$$

The equation is now solved.

$$x=\sqrt{58}-7$$ $$x=-\sqrt{58}-7$$

Steps for Completing the Square

Multiply $x-3$ and $x-3$ to get $\left(x-3\right)^{2}$.

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

Use the distributive property to multiply $2x$ by $x+4$.

$$2x^{2}+8x=\left(x-3\right)^{2}$$

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-3\right)^{2}$.

$$2x^{2}+8x=x^{2}-6x+9$$

Subtract $x^{2}$ from both sides.

$$2x^{2}+8x-x^{2}=-6x+9$$

Combine $2x^{2}$ and $-x^{2}$ to get $x^{2}$.

$$x^{2}+8x=-6x+9$$

Add $6x$ to both sides.

$$x^{2}+8x+6x=9$$

Combine $8x$ and $6x$ to get $14x$.

$$x^{2}+14x=9$$

Divide $14$, the coefficient of the $x$ term, by $2$ to get $7$. Then add the square of $7$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+14x+7^{2}=9+7^{2}$$

Square $7$.

$$x^{2}+14x+49=9+49$$

Add $9$ to $49$.

$$x^{2}+14x+49=58$$

Factor $x^{2}+14x+49$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+7\right)^{2}=58$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+7\right)^{2}}=\sqrt{58}$$

Simplify.

$$x+7=\sqrt{58}$$ $$x+7=-\sqrt{58}$$

Subtract $7$ from both sides of the equation.

$$x=\sqrt{58}-7$$ $$x=-\sqrt{58}-7$$

Solve for x

$x=\sqrt{58}-7\approx 0.615773106$
$x=-\sqrt{58}-7\approx -14.615773106$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Multiply $x-3$ and $x-3$ to get $\left(x-3\right)^{2}$.

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

Use the distributive property to multiply $2x$ by $x+4$.

$$2x^{2}+8x=\left(x-3\right)^{2}$$

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-3\right)^{2}$.

$$2x^{2}+8x=x^{2}-6x+9$$

Subtract $x^{2}$ from both sides.

$$2x^{2}+8x-x^{2}=-6x+9$$

Combine $2x^{2}$ and $-x^{2}$ to get $x^{2}$.

$$x^{2}+8x=-6x+9$$

Add $6x$ to both sides.

$$x^{2}+8x+6x=9$$

Combine $8x$ and $6x$ to get $14x$.

$$x^{2}+14x=9$$

Subtract $9$ from both sides.

$$x^{2}+14x-9=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $14$ for $b$, and $-9$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-14±\sqrt{14^{2}-4\left(-9\right)}}{2}$$

Square $14$.

$$x=\frac{-14±\sqrt{196-4\left(-9\right)}}{2}$$

Multiply $-4$ times $-9$.

$$x=\frac{-14±\sqrt{196+36}}{2}$$

Add $196$ to $36$.

$$x=\frac{-14±\sqrt{232}}{2}$$

Take the square root of $232$.

$$x=\frac{-14±2\sqrt{58}}{2}$$

Now solve the equation $x=\frac{-14±2\sqrt{58}}{2}$ when $±$ is plus. Add $-14$ to $2\sqrt{58}$.

$$x=\frac{2\sqrt{58}-14}{2}$$

Divide $-14+2\sqrt{58}$ by $2$.

$$x=\sqrt{58}-7$$

Now solve the equation $x=\frac{-14±2\sqrt{58}}{2}$ when $±$ is minus. Subtract $2\sqrt{58}$ from $-14$.

$$x=\frac{-2\sqrt{58}-14}{2}$$

Divide $-14-2\sqrt{58}$ by $2$.

$$x=-\sqrt{58}-7$$

The equation is now solved.

$$x=\sqrt{58}-7$$ $$x=-\sqrt{58}-7$$

Steps for Completing the Square

Multiply $x-3$ and $x-3$ to get $\left(x-3\right)^{2}$.

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

Use the distributive property to multiply $2x$ by $x+4$.

$$2x^{2}+8x=\left(x-3\right)^{2}$$

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-3\right)^{2}$.

$$2x^{2}+8x=x^{2}-6x+9$$

Subtract $x^{2}$ from both sides.

$$2x^{2}+8x-x^{2}=-6x+9$$

Combine $2x^{2}$ and $-x^{2}$ to get $x^{2}$.

$$x^{2}+8x=-6x+9$$

Add $6x$ to both sides.

$$x^{2}+8x+6x=9$$

Combine $8x$ and $6x$ to get $14x$.

$$x^{2}+14x=9$$

Divide $14$, the coefficient of the $x$ term, by $2$ to get $7$. Then add the square of $7$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+14x+7^{2}=9+7^{2}$$

Square $7$.

$$x^{2}+14x+49=9+49$$

Add $9$ to $49$.

$$x^{2}+14x+49=58$$

Factor $x^{2}+14x+49$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+7\right)^{2}=58$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+7\right)^{2}}=\sqrt{58}$$

Simplify.

$$x+7=\sqrt{58}$$ $$x+7=-\sqrt{58}$$

Subtract $7$ from both sides of the equation.

$$x=\sqrt{58}-7$$ $$x=-\sqrt{58}-7$$