Solve 192=x^2+4x

$$192= { x }^{ 2 } +4x$$

Solve for x

$$x=-16$$

$$x=12$$

Steps Using Factoring

Swap sides so that all variable terms are on the left hand side.

$$x^{2}+4x=192$$

Subtract 192 from both sides.

$$x^{2}+4x-192=0$$

To solve the equation, factor x^{2}+4x-192 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.

$$a+b=4 ab=-192$$

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 -192.

$$-1,192 -2,96 -3,64 -4,48 -6,32 -8,24 -12,16$$

Calculate the sum for each pair.

$$-1+192=191 -2+96=94 -3+64=61 -4+48=44 -6+32=26 -8+24=16 -12+16=4$$

The solution is the pair that gives sum 4.

$$a=-12 b=16$$

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

$$\left(x-12\right)\left(x+16\right)$$

To find equation solutions, solve x-12=0 and x+16=0.

$$x=12 x=-16$$

Steps Using Factoring By Grouping

Swap sides so that all variable terms are on the left hand side.

$$x^{2}+4x=192$$

Subtract 192 from both sides.

$$x^{2}+4x-192=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-192. To find a and b, set up a system to be solved.

$$a+b=4 ab=1\left(-192\right)=-192$$

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 -192.

$$-1,192 -2,96 -3,64 -4,48 -6,32 -8,24 -12,16$$

Calculate the sum for each pair.

$$-1+192=191 -2+96=94 -3+64=61 -4+48=44 -6+32=26 -8+24=16 -12+16=4$$

The solution is the pair that gives sum 4.

$$a=-12 b=16$$

Rewrite x^{2}+4x-192 as \left(x^{2}-12x\right)+\left(16x-192\right).

$$\left(x^{2}-12x\right)+\left(16x-192\right)$$

Factor out x in the first and 16 in the second group.

$$x\left(x-12\right)+16\left(x-12\right)$$

Factor out common term x-12 by using distributive property.

$$\left(x-12\right)\left(x+16\right)$$

To find equation solutions, solve x-12=0 and x+16=0.

$$x=12 x=-16$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$x^{2}+4x=192$$

Subtract 192 from both sides.

$$x^{2}+4x-192=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 4 for b, and -192 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\left(-192\right)}}{2}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\left(-192\right)}}{2}$$

Multiply -4 times -192.

$$x=\frac{-4±\sqrt{16+768}}{2}$$

Add 16 to 768.

$$x=\frac{-4±\sqrt{784}}{2}$$

Take the square root of 784.

$$x=\frac{-4±28}{2}$$

Now solve the equation x=\frac{-4±28}{2} when ± is plus. Add -4 to 28.

$$x=\frac{24}{2}$$

Divide 24 by 2.

$$x=12$$

Now solve the equation x=\frac{-4±28}{2} when ± is minus. Subtract 28 from -4.

$$x=-\frac{32}{2}$$

Divide -32 by 2.

$$x=-16$$

The equation is now solved.

$$x=12 x=-16$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$x^{2}+4x=192$$

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.

$$x^{2}+4x+2^{2}=192+2^{2}$$

Square 2.

$$x^{2}+4x+4=192+4$$

Add 192 to 4.

$$x^{2}+4x+4=196$$

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}.

$$\left(x+2\right)^{2}=196$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{196}$$

Simplify.

$$x+2=14 x+2=-14$$

Subtract 2 from both sides of the equation.

$$x=12 x=-16$$