Solve Write the quadratic equation in standard form: - 4x - 14 = 3 + x ^ 2 | 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

$$\left. \begin{array} { l } { \quad - 4 x - 14 = 3 + x ^ { 2 } } \end{array} \right.$$

Solve for x

$x=-\sqrt{13}i-2\approx -2-3.605551275i$
$x=-2+\sqrt{13}i\approx -2+3.605551275i$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Subtract $3$ from both sides.

$$-4x-14-3=x^{2}$$

Subtract $3$ from $-14$ to get $-17$.

$$-4x-17=x^{2}$$

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

$$-4x-17-x^{2}=0$$

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.

$$-x^{2}-4x-17=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-1$ for $a$, $-4$ for $b$, and $-17$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-1\right)\left(-17\right)}}{2\left(-1\right)}$$

Square $-4$.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-1\right)\left(-17\right)}}{2\left(-1\right)}$$

Multiply $-4$ times $-1$.

$$x=\frac{-\left(-4\right)±\sqrt{16+4\left(-17\right)}}{2\left(-1\right)}$$

Multiply $4$ times $-17$.

$$x=\frac{-\left(-4\right)±\sqrt{16-68}}{2\left(-1\right)}$$

Add $16$ to $-68$.

$$x=\frac{-\left(-4\right)±\sqrt{-52}}{2\left(-1\right)}$$

Take the square root of $-52$.

$$x=\frac{-\left(-4\right)±2\sqrt{13}i}{2\left(-1\right)}$$

The opposite of $-4$ is $4$.

$$x=\frac{4±2\sqrt{13}i}{2\left(-1\right)}$$

Multiply $2$ times $-1$.

$$x=\frac{4±2\sqrt{13}i}{-2}$$

Now solve the equation $x=\frac{4±2\sqrt{13}i}{-2}$ when $±$ is plus. Add $4$ to $2i\sqrt{13}$.

$$x=\frac{4+2\sqrt{13}i}{-2}$$

Divide $4+2i\sqrt{13}$ by $-2$.

$$x=-\sqrt{13}i-2$$

Now solve the equation $x=\frac{4±2\sqrt{13}i}{-2}$ when $±$ is minus. Subtract $2i\sqrt{13}$ from $4$.

$$x=\frac{-2\sqrt{13}i+4}{-2}$$

Divide $4-2i\sqrt{13}$ by $-2$.

$$x=-2+\sqrt{13}i$$

The equation is now solved.

$$x=-\sqrt{13}i-2$$ $$x=-2+\sqrt{13}i$$

Steps for Completing the Square

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

$$-4x-14-x^{2}=3$$

Add $14$ to both sides.

$$-4x-x^{2}=3+14$$

Add $3$ and $14$ to get $17$.

$$-4x-x^{2}=17$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form $x^{2}+bx=c$.

$$-x^{2}-4x=17$$

Divide both sides by $-1$.

$$\frac{-x^{2}-4x}{-1}=\frac{17}{-1}$$

Dividing by $-1$ undoes the multiplication by $-1$.

$$x^{2}+\left(-\frac{4}{-1}\right)x=\frac{17}{-1}$$

Divide $-4$ by $-1$.

$$x^{2}+4x=\frac{17}{-1}$$

Divide $17$ by $-1$.

$$x^{2}+4x=-17$$

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.

$$x^{2}+4x+2^{2}=-17+2^{2}$$

Square $2$.

$$x^{2}+4x+4=-17+4$$

Add $-17$ to $4$.

$$x^{2}+4x+4=-13$$

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

$$\left(x+2\right)^{2}=-13$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{-13}$$

Simplify.

$$x+2=\sqrt{13}i$$ $$x+2=-\sqrt{13}i$$

Subtract $2$ from both sides of the equation.

$$x=-2+\sqrt{13}i$$ $$x=-\sqrt{13}i-2$$