Solve Write the quadratic equation in standard form: - 10 + x ^ 2 = - 2x | Equatix - Math Solver

Fractions & Decimals

Ratios & Proportions

Basic Number Theory

Integers & Absolute Values

Factors & Multiples

Prime Factorization

Basic Equations & Inequalities

Coordinate Plane & Graphing

Linear Equations & Inequalities

Quadratic Equations

Polynomials & Factoring

Functions & Graphs

Systems of Equations

Trigonometric Functions

Trigonometric Identities & Equations

Right Triangle Trigonometry

Graphing Trigonometric Functions

Limits & Continuity

Derivatives & Applications

Integrals & Applications

Series & Sequences

Multivariable Calculus

Functions & Graphs

Polynomial & Rational Functions

Exponential & Logarithmic Functions

Sequences & Series

Descriptive Statistics

Normal Distribution

Hypothesis Testing

Regression Analysis

Linear Programming

Sets & Counting

Eigenvalues & Eigenvectors

Linear Transformations

Atomic Structure

Thermochemistry

Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$\left. \begin{array} { l } { \quad - 10 + x ^ { 2 } = - 2 x } \end{array} \right.$$

Solve for x (complex solution)

$x=\sqrt{11}-1\approx 2.31662479$
$x=-\left(\sqrt{11}+1\right)\approx -4.31662479$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Add $2x$ to both sides.

$$-10+x^{2}+2x=0$$

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.

$$x^{2}+2x-10=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $2$ for $b$, and $-10$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-10\right)}}{2}$$

Square $2$.

$$x=\frac{-2±\sqrt{4-4\left(-10\right)}}{2}$$

Multiply $-4$ times $-10$.

$$x=\frac{-2±\sqrt{4+40}}{2}$$

Add $4$ to $40$.

$$x=\frac{-2±\sqrt{44}}{2}$$

Take the square root of $44$.

$$x=\frac{-2±2\sqrt{11}}{2}$$

Now solve the equation $x=\frac{-2±2\sqrt{11}}{2}$ when $±$ is plus. Add $-2$ to $2\sqrt{11}$.

$$x=\frac{2\sqrt{11}-2}{2}$$

Divide $-2+2\sqrt{11}$ by $2$.

$$x=\sqrt{11}-1$$

Now solve the equation $x=\frac{-2±2\sqrt{11}}{2}$ when $±$ is minus. Subtract $2\sqrt{11}$ from $-2$.

$$x=\frac{-2\sqrt{11}-2}{2}$$

Divide $-2-2\sqrt{11}$ by $2$.

$$x=-\sqrt{11}-1$$

The equation is now solved.

$$x=\sqrt{11}-1$$ $$x=-\sqrt{11}-1$$

Steps for Completing the Square

Add $2x$ to both sides.

$$-10+x^{2}+2x=0$$

Add $10$ to both sides. Anything plus zero gives itself.

$$x^{2}+2x=10$$

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

$$x^{2}+2x+1^{2}=10+1^{2}$$

Square $1$.

$$x^{2}+2x+1=10+1$$

Add $10$ to $1$.

$$x^{2}+2x+1=11$$

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

$$\left(x+1\right)^{2}=11$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{11}$$

Simplify.

$$x+1=\sqrt{11}$$ $$x+1=-\sqrt{11}$$

Subtract $1$ from both sides of the equation.

$$x=\sqrt{11}-1$$ $$x=-\sqrt{11}-1$$

Solve for x

$x=\sqrt{11}-1\approx 2.31662479$
$x=-\sqrt{11}-1\approx -4.31662479$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Add $2x$ to both sides.

$$-10+x^{2}+2x=0$$

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.

$$x^{2}+2x-10=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $2$ for $b$, and $-10$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-10\right)}}{2}$$

Square $2$.

$$x=\frac{-2±\sqrt{4-4\left(-10\right)}}{2}$$

Multiply $-4$ times $-10$.

$$x=\frac{-2±\sqrt{4+40}}{2}$$

Add $4$ to $40$.

$$x=\frac{-2±\sqrt{44}}{2}$$

Take the square root of $44$.

$$x=\frac{-2±2\sqrt{11}}{2}$$

Now solve the equation $x=\frac{-2±2\sqrt{11}}{2}$ when $±$ is plus. Add $-2$ to $2\sqrt{11}$.

$$x=\frac{2\sqrt{11}-2}{2}$$

Divide $-2+2\sqrt{11}$ by $2$.

$$x=\sqrt{11}-1$$

Now solve the equation $x=\frac{-2±2\sqrt{11}}{2}$ when $±$ is minus. Subtract $2\sqrt{11}$ from $-2$.

$$x=\frac{-2\sqrt{11}-2}{2}$$

Divide $-2-2\sqrt{11}$ by $2$.

$$x=-\sqrt{11}-1$$

The equation is now solved.

$$x=\sqrt{11}-1$$ $$x=-\sqrt{11}-1$$

Steps for Completing the Square

Add $2x$ to both sides.

$$-10+x^{2}+2x=0$$

Add $10$ to both sides. Anything plus zero gives itself.

$$x^{2}+2x=10$$

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

$$x^{2}+2x+1^{2}=10+1^{2}$$

Square $1$.

$$x^{2}+2x+1=10+1$$

Add $10$ to $1$.

$$x^{2}+2x+1=11$$

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

$$\left(x+1\right)^{2}=11$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{11}$$

Simplify.

$$x+1=\sqrt{11}$$ $$x+1=-\sqrt{11}$$

Subtract $1$ from both sides of the equation.

$$x=\sqrt{11}-1$$ $$x=-\sqrt{11}-1$$