Solve x^2=7x+44

$$x ^ { 2 } = 7 x + 44$$

Solve for x

$$x=-4$$

$$x=11$$

Steps Using Factoring

Subtract 7x from both sides.

$$x^{2}-7x=44$$

Subtract 44 from both sides.

$$x^{2}-7x-44=0$$

To solve the equation, factor x^{2}-7x-44 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.

$$a+b=-7 ab=-44$$

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 -44.

$$1,-44 2,-22 4,-11$$

Calculate the sum for each pair.

$$1-44=-43 2-22=-20 4-11=-7$$

The solution is the pair that gives sum -7.

$$a=-11 b=4$$

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

$$\left(x-11\right)\left(x+4\right)$$

To find equation solutions, solve x-11=0 and x+4=0.

$$x=11 x=-4$$

Steps Using Factoring By Grouping

Subtract 7x from both sides.

$$x^{2}-7x=44$$

Subtract 44 from both sides.

$$x^{2}-7x-44=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-44. To find a and b, set up a system to be solved.

$$a+b=-7 ab=1\left(-44\right)=-44$$

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 -44.

$$1,-44 2,-22 4,-11$$

Calculate the sum for each pair.

$$1-44=-43 2-22=-20 4-11=-7$$

The solution is the pair that gives sum -7.

$$a=-11 b=4$$

Rewrite x^{2}-7x-44 as \left(x^{2}-11x\right)+\left(4x-44\right).

$$\left(x^{2}-11x\right)+\left(4x-44\right)$$

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

$$x\left(x-11\right)+4\left(x-11\right)$$

Factor out common term x-11 by using distributive property.

$$\left(x-11\right)\left(x+4\right)$$

To find equation solutions, solve x-11=0 and x+4=0.

$$x=11 x=-4$$

Steps Using the Quadratic Formula

Subtract 7x from both sides.

$$x^{2}-7x=44$$

Subtract 44 from both sides.

$$x^{2}-7x-44=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -7 for b, and -44 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\left(-44\right)}}{2}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\left(-44\right)}}{2}$$

Multiply -4 times -44.

$$x=\frac{-\left(-7\right)±\sqrt{49+176}}{2}$$

Add 49 to 176.

$$x=\frac{-\left(-7\right)±\sqrt{225}}{2}$$

Take the square root of 225.

$$x=\frac{-\left(-7\right)±15}{2}$$

The opposite of -7 is 7.

$$x=\frac{7±15}{2}$$

Now solve the equation x=\frac{7±15}{2} when ± is plus. Add 7 to 15.

$$x=\frac{22}{2}$$

Divide 22 by 2.

$$x=11$$

Now solve the equation x=\frac{7±15}{2} when ± is minus. Subtract 15 from 7.

$$x=-\frac{8}{2}$$

Divide -8 by 2.

$$x=-4$$

The equation is now solved.

$$x=11 x=-4$$

Steps for Completing the Square

Subtract 7x from both sides.

$$x^{2}-7x=44$$

Divide -7, the coefficient of the x term, by 2 to get -\frac{7}{2}. Then add the square of -\frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-7x+\left(-\frac{7}{2}\right)^{2}=44+\left(-\frac{7}{2}\right)^{2}$$

Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-7x+\frac{49}{4}=44+\frac{49}{4}$$

Add 44 to \frac{49}{4}.

$$x^{2}-7x+\frac{49}{4}=\frac{225}{4}$$

Factor x^{2}-7x+\frac{49}{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}.

$$\left(x-\frac{7}{2}\right)^{2}=\frac{225}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{2}\right)^{2}}=\sqrt{\frac{225}{4}}$$

Simplify.

$$x-\frac{7}{2}=\frac{15}{2} x-\frac{7}{2}=-\frac{15}{2}$$

Add \frac{7}{2} to both sides of the equation.

$$x=11 x=-4$$