Solve 4x^2+3x-36=104-5x

$$4 x ^ { 2 } + 3 x - 36 = 104 - 5 x$$

Solve for x

$$x=-7$$

$$x=5$$

Steps Using Factoring By Grouping

Subtract 104 from both sides.

$$4x^{2}+3x-36-104=-5x$$

Subtract 104 from -36 to get -140.

$$4x^{2}+3x-140=-5x$$

Add 5x to both sides.

$$4x^{2}+3x-140+5x=0$$

Combine 3x and 5x to get 8x.

$$4x^{2}+8x-140=0$$

Divide both sides by 4.

$$x^{2}+2x-35=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-35. To find a and b, set up a system to be solved.

$$a+b=2 ab=1\left(-35\right)=-35$$

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 -35.

$$-1,35 -5,7$$

Calculate the sum for each pair.

$$-1+35=34 -5+7=2$$

The solution is the pair that gives sum 2.

$$a=-5 b=7$$

Rewrite x^{2}+2x-35 as \left(x^{2}-5x\right)+\left(7x-35\right).

$$\left(x^{2}-5x\right)+\left(7x-35\right)$$

Factor out x in the first and 7 in the second group.

$$x\left(x-5\right)+7\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(x+7\right)$$

To find equation solutions, solve x-5=0 and x+7=0.

$$x=5 x=-7$$

Steps Using the Quadratic Formula

Subtract 104 from both sides.

$$4x^{2}+3x-36-104=-5x$$

Subtract 104 from -36 to get -140.

$$4x^{2}+3x-140=-5x$$

Add 5x to both sides.

$$4x^{2}+3x-140+5x=0$$

Combine 3x and 5x to get 8x.

$$4x^{2}+8x-140=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 8 for b, and -140 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\times 4\left(-140\right)}}{2\times 4}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 4\left(-140\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-8±\sqrt{64-16\left(-140\right)}}{2\times 4}$$

Multiply -16 times -140.

$$x=\frac{-8±\sqrt{64+2240}}{2\times 4}$$

Add 64 to 2240.

$$x=\frac{-8±\sqrt{2304}}{2\times 4}$$

Take the square root of 2304.

$$x=\frac{-8±48}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-8±48}{8}$$

Now solve the equation x=\frac{-8±48}{8} when ± is plus. Add -8 to 48.

$$x=\frac{40}{8}$$

Divide 40 by 8.

$$x=5$$

Now solve the equation x=\frac{-8±48}{8} when ± is minus. Subtract 48 from -8.

$$x=-\frac{56}{8}$$

Divide -56 by 8.

$$x=-7$$

The equation is now solved.

$$x=5 x=-7$$

Steps for Completing the Square

Add 5x to both sides.

$$4x^{2}+3x-36+5x=104$$

Combine 3x and 5x to get 8x.

$$4x^{2}+8x-36=104$$

Add 36 to both sides.

$$4x^{2}+8x=104+36$$

Add 104 and 36 to get 140.

$$4x^{2}+8x=140$$

Divide both sides by 4.

$$\frac{4x^{2}+8x}{4}=\frac{140}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{8}{4}x=\frac{140}{4}$$

Divide 8 by 4.

$$x^{2}+2x=\frac{140}{4}$$

Divide 140 by 4.

$$x^{2}+2x=35$$

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+2x+1^{2}=35+1^{2}$$

Square 1.

$$x^{2}+2x+1=35+1$$

Add 35 to 1.

$$x^{2}+2x+1=36$$

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+1\right)^{2}=36$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{36}$$

Simplify.

$$x+1=6 x+1=-6$$

Subtract 1 from both sides of the equation.

$$x=5 x=-7$$