Type a math problem

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Type a math problem

Solve for x

x=8<br/>x=2

$x=8$

$x=2$

$x=2$

Steps Using Factoring

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring

( x - 5 ) ^ { 2 } - 9 = 0

$(x−5)_{2}−9=0$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

Use binomial theorem $(a−b)_{2}=a_{2}−2ab+b_{2}$ to expand $(x−5)_{2}$.

x^{2}-10x+25-9=0

$x_{2}−10x+25−9=0$

Subtract 9 from 25 to get 16.

Subtract $9$ from $25$ to get $16$.

x^{2}-10x+16=0

$x_{2}−10x+16=0$

To solve the equation, factor x^{2}-10x+16 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

To solve the equation, factor $x_{2}−10x+16$ using formula $x_{2}+(a+b)x+ab=(x+a)(x+b)$. To find $a$ and $b$, set up a system to be solved.

a+b=-10 ab=16

$a+b=−10$ $ab=16$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 16.

Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is negative, $a$ and $b$ are both negative. List all such integer pairs that give product $16$.

-1,-16 -2,-8 -4,-4

$−1,−16$ $−2,−8$ $−4,−4$

Calculate the sum for each pair.

Calculate the sum for each pair.

-1-16=-17 -2-8=-10 -4-4=-8

$−1−16=−17$ $−2−8=−10$ $−4−4=−8$

The solution is the pair that gives sum -10.

The solution is the pair that gives sum $−10$.

a=-8 b=-2

$a=−8$ $b=−2$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

Rewrite factored expression $(x+a)(x+b)$ using the obtained values.

\left(x-8\right)\left(x-2\right)

$(x−8)(x−2)$

To find equation solutions, solve x-8=0 and x-2=0.

To find equation solutions, solve $x−8=0$ and $x−2=0$.

x=8 x=2

$x=8$ $x=2$

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x^{2}-10x+25-9=0

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

x^{2}-10x+16=0

Subtract 9 from 25 to get 16.

a+b=-10 ab=16

To solve the equation, factor x^{2}-10x+16 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

-1,-16 -2,-8 -4,-4

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 16.

-1-16=-17 -2-8=-10 -4-4=-8

Calculate the sum for each pair.

a=-8 b=-2

The solution is the pair that gives sum -10.

\left(x-8\right)\left(x-2\right)

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

x=8 x=2

To find equation solutions, solve x-8=0 and x-2=0.

x^{2}-10x+25-9=0

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

x^{2}-10x+16=0

Subtract 9 from 25 to get 16.

a+b=-10 ab=1\times 16=16

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+16. To find a and b, set up a system to be solved.

-1,-16 -2,-8 -4,-4

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 16.

-1-16=-17 -2-8=-10 -4-4=-8

Calculate the sum for each pair.

a=-8 b=-2

The solution is the pair that gives sum -10.

\left(x^{2}-8x\right)+\left(-2x+16\right)

Rewrite x^{2}-10x+16 as \left(x^{2}-8x\right)+\left(-2x+16\right).

x\left(x-8\right)-2\left(x-8\right)

Factor out x in the first and -2 in the second group.

\left(x-8\right)\left(x-2\right)

Factor out common term x-8 by using distributive property.

x=8 x=2

To find equation solutions, solve x-8=0 and x-2=0.

x^{2}-10x+25-9=0

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

x^{2}-10x+16=0

Subtract 9 from 25 to get 16.

x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 16}}{2}

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and 16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

x=\frac{-\left(-10\right)±\sqrt{100-4\times 16}}{2}

Square -10.

x=\frac{-\left(-10\right)±\sqrt{100-64}}{2}

Multiply -4 times 16.

x=\frac{-\left(-10\right)±\sqrt{36}}{2}

Add 100 to -64.

x=\frac{-\left(-10\right)±6}{2}

Take the square root of 36.

x=\frac{10±6}{2}

The opposite of -10 is 10.

x=\frac{16}{2}

Now solve the equation x=\frac{10±6}{2} when ± is plus. Add 10 to 6.

x=8

Divide 16 by 2.

x=\frac{4}{2}

Now solve the equation x=\frac{10±6}{2} when ± is minus. Subtract 6 from 10.

x=2

Divide 4 by 2.

x=8 x=2

The equation is now solved.

x^{2}-10x+25-9=0

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

x^{2}-10x+16=0

Subtract 9 from 25 to get 16.

x^{2}-10x=-16

Subtract 16 from both sides. Anything subtracted from zero gives its negation.

x^{2}-10x+\left(-5\right)^{2}=-16+\left(-5\right)^{2}

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

x^{2}-10x+25=-16+25

Square -5.

x^{2}-10x+25=9

Add -16 to 25.

\left(x-5\right)^{2}=9

Factor x^{2}-10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

\sqrt{\left(x-5\right)^{2}}=\sqrt{9}

Take the square root of both sides of the equation.

x-5=3 x-5=-3

Simplify.

x=8 x=2

Add 5 to both sides of the equation.

Examples

Quadratic equation

{ x } ^ { 2 } - 4 x - 5 = 0

$x_{2}−4x−5=0$

Trigonometry

4 \sin \theta \cos \theta = 2 \sin \theta

$4sinθcosθ=2sinθ$

Linear equation

y = 3x + 4

$y=3x+4$

Arithmetic

699 * 533

$699∗533$

Matrix

\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { - 1 } & { 1 } & { 5 } \end{array} \right]

$[25 34 ][2−1 01 35 ]$

Simultaneous equation

\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.

${8x+2y=467x+3y=47 $

Differentiation

\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }

$dxd (x−5)(3x_{2}−2) $

Integration

\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x

$∫_{0}xe_{−x_{2}}dx$

Limits

\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}

$x→−3lim x_{2}+2x−3x_{2}−9 $

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