Solve 40*22-(22x+40x-x^2)=760

$$40 \times 22 - ( 22 x + 40 x - x ^ { 2 } ) = 760$$

Solve for x

$$x=2$$

$$x=60$$

Steps Using Factoring

Multiply 40 and 22 to get 880.

$$880-\left(22x+40x-x^{2}\right)=760$$

Combine 22x and 40x to get 62x.

$$880-\left(62x-x^{2}\right)=760$$

To find the opposite of 62x-x^{2}, find the opposite of each term.

$$880-62x+x^{2}=760$$

Subtract 760 from both sides.

$$880-62x+x^{2}-760=0$$

Subtract 760 from 880 to get 120.

$$120-62x+x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}-62x+120=0$$

To solve the equation, factor x^{2}-62x+120 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.

$$a+b=-62 ab=120$$

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 120.

$$-1,-120 -2,-60 -3,-40 -4,-30 -5,-24 -6,-20 -8,-15 -10,-12$$

Calculate the sum for each pair.

$$-1-120=-121 -2-60=-62 -3-40=-43 -4-30=-34 -5-24=-29 -6-20=-26 -8-15=-23 -10-12=-22$$

The solution is the pair that gives sum -62.

$$a=-60 b=-2$$

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

$$\left(x-60\right)\left(x-2\right)$$

To find equation solutions, solve x-60=0 and x-2=0.

$$x=60 x=2$$

Steps Using Factoring By Grouping

Multiply 40 and 22 to get 880.

$$880-\left(22x+40x-x^{2}\right)=760$$

Combine 22x and 40x to get 62x.

$$880-\left(62x-x^{2}\right)=760$$

To find the opposite of 62x-x^{2}, find the opposite of each term.

$$880-62x+x^{2}=760$$

Subtract 760 from both sides.

$$880-62x+x^{2}-760=0$$

Subtract 760 from 880 to get 120.

$$120-62x+x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}-62x+120=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+120. To find a and b, set up a system to be solved.

$$a+b=-62 ab=1\times 120=120$$

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 120.

$$-1,-120 -2,-60 -3,-40 -4,-30 -5,-24 -6,-20 -8,-15 -10,-12$$

Calculate the sum for each pair.

$$-1-120=-121 -2-60=-62 -3-40=-43 -4-30=-34 -5-24=-29 -6-20=-26 -8-15=-23 -10-12=-22$$

The solution is the pair that gives sum -62.

$$a=-60 b=-2$$

Rewrite x^{2}-62x+120 as \left(x^{2}-60x\right)+\left(-2x+120\right).

$$\left(x^{2}-60x\right)+\left(-2x+120\right)$$

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

$$x\left(x-60\right)-2\left(x-60\right)$$

Factor out common term x-60 by using distributive property.

$$\left(x-60\right)\left(x-2\right)$$

To find equation solutions, solve x-60=0 and x-2=0.

$$x=60 x=2$$

Steps Using the Quadratic Formula

Multiply 40 and 22 to get 880.

$$880-\left(22x+40x-x^{2}\right)=760$$

Combine 22x and 40x to get 62x.

$$880-\left(62x-x^{2}\right)=760$$

To find the opposite of 62x-x^{2}, find the opposite of each term.

$$880-62x+x^{2}=760$$

Subtract 760 from both sides.

$$880-62x+x^{2}-760=0$$

Subtract 760 from 880 to get 120.

$$120-62x+x^{2}=0$$

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.

$$x^{2}-62x+120=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -62 for b, and 120 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-62\right)±\sqrt{\left(-62\right)^{2}-4\times 120}}{2}$$

Square -62.

$$x=\frac{-\left(-62\right)±\sqrt{3844-4\times 120}}{2}$$

Multiply -4 times 120.

$$x=\frac{-\left(-62\right)±\sqrt{3844-480}}{2}$$

Add 3844 to -480.

$$x=\frac{-\left(-62\right)±\sqrt{3364}}{2}$$

Take the square root of 3364.

$$x=\frac{-\left(-62\right)±58}{2}$$

The opposite of -62 is 62.

$$x=\frac{62±58}{2}$$

Now solve the equation x=\frac{62±58}{2} when ± is plus. Add 62 to 58.

$$x=\frac{120}{2}$$

Divide 120 by 2.

$$x=60$$

Now solve the equation x=\frac{62±58}{2} when ± is minus. Subtract 58 from 62.

$$x=\frac{4}{2}$$

Divide 4 by 2.

$$x=2$$

The equation is now solved.

$$x=60 x=2$$

Steps for Completing the Square

Multiply 40 and 22 to get 880.

$$880-\left(22x+40x-x^{2}\right)=760$$

Combine 22x and 40x to get 62x.

$$880-\left(62x-x^{2}\right)=760$$

To find the opposite of 62x-x^{2}, find the opposite of each term.

$$880-62x+x^{2}=760$$

Subtract 880 from both sides.

$$-62x+x^{2}=760-880$$

Subtract 880 from 760 to get -120.

$$-62x+x^{2}=-120$$

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.

$$x^{2}-62x=-120$$

Divide -62, the coefficient of the x term, by 2 to get -31. Then add the square of -31 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-62x+\left(-31\right)^{2}=-120+\left(-31\right)^{2}$$

Square -31.

$$x^{2}-62x+961=-120+961$$

Add -120 to 961.

$$x^{2}-62x+961=841$$

Factor x^{2}-62x+961. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-31\right)^{2}=841$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-31\right)^{2}}=\sqrt{841}$$

Simplify.

$$x-31=29 x-31=-29$$

Add 31 to both sides of the equation.

$$x=60 x=2$$