Solve 120+x+x^2/4-3=180

$$120+x+ \frac{ { x }^{ 2 } }{ 4 } -3=180$$

Solve for x

$$x=-18$$

$$x=14$$

Steps Using Factoring

Multiply both sides of the equation by 4.

$$480+4x+x^{2}-12=720$$

Subtract 12 from 480 to get 468.

$$468+4x+x^{2}=720$$

Subtract 720 from both sides.

$$468+4x+x^{2}-720=0$$

Subtract 720 from 468 to get -252.

$$-252+4x+x^{2}=0$$

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

$$x^{2}+4x-252=0$$

To solve the equation, factor x^{2}+4x-252 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.

$$a+b=4 ab=-252$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -252.

$$-1,252 -2,126 -3,84 -4,63 -6,42 -7,36 -9,28 -12,21 -14,18$$

Calculate the sum for each pair.

$$-1+252=251 -2+126=124 -3+84=81 -4+63=59 -6+42=36 -7+36=29 -9+28=19 -12+21=9 -14+18=4$$

The solution is the pair that gives sum 4.

$$a=-14 b=18$$

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

$$\left(x-14\right)\left(x+18\right)$$

To find equation solutions, solve x-14=0 and x+18=0.

$$x=14 x=-18$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 4.

$$480+4x+x^{2}-12=720$$

Subtract 12 from 480 to get 468.

$$468+4x+x^{2}=720$$

Subtract 720 from both sides.

$$468+4x+x^{2}-720=0$$

Subtract 720 from 468 to get -252.

$$-252+4x+x^{2}=0$$

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

$$x^{2}+4x-252=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-252. To find a and b, set up a system to be solved.

$$a+b=4 ab=1\left(-252\right)=-252$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -252.

$$-1,252 -2,126 -3,84 -4,63 -6,42 -7,36 -9,28 -12,21 -14,18$$

Calculate the sum for each pair.

$$-1+252=251 -2+126=124 -3+84=81 -4+63=59 -6+42=36 -7+36=29 -9+28=19 -12+21=9 -14+18=4$$

The solution is the pair that gives sum 4.

$$a=-14 b=18$$

Rewrite x^{2}+4x-252 as \left(x^{2}-14x\right)+\left(18x-252\right).

$$\left(x^{2}-14x\right)+\left(18x-252\right)$$

Factor out x in the first and 18 in the second group.

$$x\left(x-14\right)+18\left(x-14\right)$$

Factor out common term x-14 by using distributive property.

$$\left(x-14\right)\left(x+18\right)$$

To find equation solutions, solve x-14=0 and x+18=0.

$$x=14 x=-18$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 4.

$$480+4x+x^{2}-12=720$$

Subtract 12 from 480 to get 468.

$$468+4x+x^{2}=720$$

Subtract 720 from both sides.

$$468+4x+x^{2}-720=0$$

Subtract 720 from 468 to get -252.

$$-252+4x+x^{2}=0$$

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.

$$x^{2}+4x-252=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 4 for b, and -252 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\left(-252\right)}}{2}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\left(-252\right)}}{2}$$

Multiply -4 times -252.

$$x=\frac{-4±\sqrt{16+1008}}{2}$$

Add 16 to 1008.

$$x=\frac{-4±\sqrt{1024}}{2}$$

Take the square root of 1024.

$$x=\frac{-4±32}{2}$$

Now solve the equation x=\frac{-4±32}{2} when ± is plus. Add -4 to 32.

$$x=\frac{28}{2}$$

Divide 28 by 2.

$$x=14$$

Now solve the equation x=\frac{-4±32}{2} when ± is minus. Subtract 32 from -4.

$$x=-\frac{36}{2}$$

Divide -36 by 2.

$$x=-18$$

The equation is now solved.

$$x=14 x=-18$$

Steps for Completing the Square

Multiply both sides of the equation by 4.

$$480+4x+x^{2}-12=720$$

Subtract 12 from 480 to get 468.

$$468+4x+x^{2}=720$$

Subtract 468 from both sides.

$$4x+x^{2}=720-468$$

Subtract 468 from 720 to get 252.

$$4x+x^{2}=252$$

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.

$$x^{2}+4x=252$$

Divide 4, the coefficient of the x term, by 2 to get 2. Then add the square of 2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+4x+2^{2}=252+2^{2}$$

Square 2.

$$x^{2}+4x+4=252+4$$

Add 252 to 4.

$$x^{2}+4x+4=256$$

Factor x^{2}+4x+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}.

$$\left(x+2\right)^{2}=256$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{256}$$

Simplify.

$$x+2=16 x+2=-16$$

Subtract 2 from both sides of the equation.

$$x=14 x=-18$$