Type a math problem

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Type a math problem

Solve for x

x=-4<br/>x=7

$x=−4$

$x=7$

$x=7$

Steps Using Factoring

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring

x^2-3x=28

$x_{2}−3x=28$

Subtract 28 from both sides.

Subtract $28$ from both sides.

x^{2}-3x-28=0

$x_{2}−3x−28=0$

To solve the equation, factor x^{2}-3x-28 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 $x_{2}−3x−28$ 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=-3 ab=-28

$a+b=−3$ $ab=−28$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -28.

Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product $−28$.

1,-28 2,-14 4,-7

$1,−28$ $2,−14$ $4,−7$

Calculate the sum for each pair.

Calculate the sum for each pair.

1-28=-27 2-14=-12 4-7=-3

$1−28=−27$ $2−14=−12$ $4−7=−3$

The solution is the pair that gives sum -3.

The solution is the pair that gives sum $−3$.

a=-7 b=4

$a=−7$ $b=4$

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-7\right)\left(x+4\right)

$(x−7)(x+4)$

To find equation solutions, solve x-7=0 and x+4=0.

To find equation solutions, solve $x−7=0$ and $x+4=0$.

x=7 x=-4

$x=7$ $x=−4$

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x^{2}-3x-28=0

Subtract 28 from both sides.

a+b=-3 ab=-28

To solve the equation, factor x^{2}-3x-28 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,-28 2,-14 4,-7

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -28.

1-28=-27 2-14=-12 4-7=-3

Calculate the sum for each pair.

a=-7 b=4

The solution is the pair that gives sum -3.

\left(x-7\right)\left(x+4\right)

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

x=7 x=-4

To find equation solutions, solve x-7=0 and x+4=0.

x^{2}-3x-28=0

Subtract 28 from both sides.

a+b=-3 ab=1\left(-28\right)=-28

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-28. To find a and b, set up a system to be solved.

1,-28 2,-14 4,-7

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -28.

1-28=-27 2-14=-12 4-7=-3

Calculate the sum for each pair.

a=-7 b=4

The solution is the pair that gives sum -3.

\left(x^{2}-7x\right)+\left(4x-28\right)

Rewrite x^{2}-3x-28 as \left(x^{2}-7x\right)+\left(4x-28\right).

x\left(x-7\right)+4\left(x-7\right)

Factor out x in the first and 4 in the second group.

\left(x-7\right)\left(x+4\right)

Factor out common term x-7 by using distributive property.

x=7 x=-4

To find equation solutions, solve x-7=0 and x+4=0.

x^{2}-3x=28

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}-3x-28=28-28

Subtract 28 from both sides of the equation.

x^{2}-3x-28=0

Subtracting 28 from itself leaves 0.

x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-28\right)}}{2}

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -3 for b, and -28 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

x=\frac{-\left(-3\right)±\sqrt{9-4\left(-28\right)}}{2}

Square -3.

x=\frac{-\left(-3\right)±\sqrt{9+112}}{2}

Multiply -4 times -28.

x=\frac{-\left(-3\right)±\sqrt{121}}{2}

Add 9 to 112.

x=\frac{-\left(-3\right)±11}{2}

Take the square root of 121.

x=\frac{3±11}{2}

The opposite of -3 is 3.

x=\frac{14}{2}

Now solve the equation x=\frac{3±11}{2} when ± is plus. Add 3 to 11.

x=7

Divide 14 by 2.

x=\frac{-8}{2}

Now solve the equation x=\frac{3±11}{2} when ± is minus. Subtract 11 from 3.

x=-4

Divide -8 by 2.

x=7 x=-4

The equation is now solved.

x^{2}-3x=28

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}-3x+\left(-\frac{3}{2}\right)^{2}=28+\left(-\frac{3}{2}\right)^{2}

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}=-1.5. Then add the square of -\frac{3}{2}=-1.5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

x^{2}-3x+\frac{9}{4}=28+\frac{9}{4}

Square -\frac{3}{2}=-1.5 by squaring both the numerator and the denominator of the fraction.

x^{2}-3x+\frac{9}{4}=\frac{121}{4}

Add 28 to \frac{9}{4}=2.25.

\left(x-\frac{3}{2}\right)^{2}=\frac{121}{4}

Factor x^{2}-3x+\frac{9}{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}.

\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{121}{4}}

Take the square root of both sides of the equation.

x-\frac{3}{2}=\frac{11}{2} x-\frac{3}{2}=-\frac{11}{2}

Simplify.

x=7 x=-4

Add \frac{3}{2}=1.5 to both sides of the equation.

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