Solve Trite the following quadratic re quadratic equations . (x + 7)(x - 3) = - 7 | 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

$$\left. \begin{array} { l } { \quad ( x + 7 ) ( x - 3 ) = - 7 } \end{array} \right.$$

Solve for x

$x=3\sqrt{2}-2\approx 2.242640687$
$x=-3\sqrt{2}-2\approx -6.242640687$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Use the distributive property to multiply $x+7$ by $x-3$ and combine like terms.

$$x^{2}+4x-21=-7$$

Add $7$ to both sides.

$$x^{2}+4x-21+7=0$$

Add $-21$ and $7$ to get $-14$.

$$x^{2}+4x-14=0$$

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

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

Square $4$.

$$x=\frac{-4±\sqrt{16-4\left(-14\right)}}{2}$$

Multiply $-4$ times $-14$.

$$x=\frac{-4±\sqrt{16+56}}{2}$$

Add $16$ to $56$.

$$x=\frac{-4±\sqrt{72}}{2}$$

Take the square root of $72$.

$$x=\frac{-4±6\sqrt{2}}{2}$$

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

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

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

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

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

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

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

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

The equation is now solved.

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

Steps for Completing the Square

Use the distributive property to multiply $x+7$ by $x-3$ and combine like terms.

$$x^{2}+4x-21=-7$$

Add $21$ to both sides.

$$x^{2}+4x=-7+21$$

Add $-7$ and $21$ to get $14$.

$$x^{2}+4x=14$$

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

$$x^{2}+4x+2^{2}=14+2^{2}$$

Square $2$.

$$x^{2}+4x+4=14+4$$

Add $14$ to $4$.

$$x^{2}+4x+4=18$$

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

$$\left(x+2\right)^{2}=18$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{18}$$

Simplify.

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

Subtract $2$ from both sides of the equation.

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