Solve 40x-80+6x+12-3x^2+6x-6x+12=0

$$40x-80+6x+12-3 { x }^{ 2 } +6x-6x+12=0$$

Solve for x

$$x=14$$

$$x = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$$

Steps Using Factoring By Grouping

Combine 40x and 6x to get 46x.

$$46x-80+12-3x^{2}+6x-6x+12=0$$

Add -80 and 12 to get -68.

$$46x-68-3x^{2}+6x-6x+12=0$$

Combine 46x and 6x to get 52x.

$$52x-68-3x^{2}-6x+12=0$$

Combine 52x and -6x to get 46x.

$$46x-68-3x^{2}+12=0$$

Add -68 and 12 to get -56.

$$46x-56-3x^{2}=0$$

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

$$-3x^{2}+46x-56=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3x^{2}+ax+bx-56. To find a and b, set up a system to be solved.

$$a+b=46 ab=-3\left(-56\right)=168$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 168.

$$1,168 2,84 3,56 4,42 6,28 7,24 8,21 12,14$$

Calculate the sum for each pair.

$$1+168=169 2+84=86 3+56=59 4+42=46 6+28=34 7+24=31 8+21=29 12+14=26$$

The solution is the pair that gives sum 46.

$$a=42 b=4$$

Rewrite -3x^{2}+46x-56 as \left(-3x^{2}+42x\right)+\left(4x-56\right).

$$\left(-3x^{2}+42x\right)+\left(4x-56\right)$$

Factor out 3x in the first and -4 in the second group.

$$3x\left(-x+14\right)-4\left(-x+14\right)$$

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

$$\left(-x+14\right)\left(3x-4\right)$$

To find equation solutions, solve -x+14=0 and 3x-4=0.

$$x=14 x=\frac{4}{3}$$

Steps Using the Quadratic Formula

Combine 40x and 6x to get 46x.

$$46x-80+12-3x^{2}+6x-6x+12=0$$

Add -80 and 12 to get -68.

$$46x-68-3x^{2}+6x-6x+12=0$$

Combine 46x and 6x to get 52x.

$$52x-68-3x^{2}-6x+12=0$$

Combine 52x and -6x to get 46x.

$$46x-68-3x^{2}+12=0$$

Add -68 and 12 to get -56.

$$46x-56-3x^{2}=0$$

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.

$$-3x^{2}+46x-56=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 46 for b, and -56 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-46±\sqrt{46^{2}-4\left(-3\right)\left(-56\right)}}{2\left(-3\right)}$$

Square 46.

$$x=\frac{-46±\sqrt{2116-4\left(-3\right)\left(-56\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-46±\sqrt{2116+12\left(-56\right)}}{2\left(-3\right)}$$

Multiply 12 times -56.

$$x=\frac{-46±\sqrt{2116-672}}{2\left(-3\right)}$$

Add 2116 to -672.

$$x=\frac{-46±\sqrt{1444}}{2\left(-3\right)}$$

Take the square root of 1444.

$$x=\frac{-46±38}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-46±38}{-6}$$

Now solve the equation x=\frac{-46±38}{-6} when ± is plus. Add -46 to 38.

$$x=-\frac{8}{-6}$$

Reduce the fraction \frac{-8}{-6} to lowest terms by extracting and canceling out 2.

$$x=\frac{4}{3}$$

Now solve the equation x=\frac{-46±38}{-6} when ± is minus. Subtract 38 from -46.

$$x=-\frac{84}{-6}$$

Divide -84 by -6.

$$x=14$$

The equation is now solved.

$$x=\frac{4}{3} x=14$$

Steps for Completing the Square

Combine 40x and 6x to get 46x.

$$46x-80+12-3x^{2}+6x-6x+12=0$$

Add -80 and 12 to get -68.

$$46x-68-3x^{2}+6x-6x+12=0$$

Combine 46x and 6x to get 52x.

$$52x-68-3x^{2}-6x+12=0$$

Combine 52x and -6x to get 46x.

$$46x-68-3x^{2}+12=0$$

Add -68 and 12 to get -56.

$$46x-56-3x^{2}=0$$

Add 56 to both sides. Anything plus zero gives itself.

$$46x-3x^{2}=56$$

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.

$$-3x^{2}+46x=56$$

Divide both sides by -3.

$$\frac{-3x^{2}+46x}{-3}=\frac{56}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{46}{-3}x=\frac{56}{-3}$$

Divide 46 by -3.

$$x^{2}-\frac{46}{3}x=\frac{56}{-3}$$

Divide 56 by -3.

$$x^{2}-\frac{46}{3}x=-\frac{56}{3}$$

Divide -\frac{46}{3}, the coefficient of the x term, by 2 to get -\frac{23}{3}. Then add the square of -\frac{23}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{46}{3}x+\left(-\frac{23}{3}\right)^{2}=-\frac{56}{3}+\left(-\frac{23}{3}\right)^{2}$$

Square -\frac{23}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{46}{3}x+\frac{529}{9}=-\frac{56}{3}+\frac{529}{9}$$

Add -\frac{56}{3} to \frac{529}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{46}{3}x+\frac{529}{9}=\frac{361}{9}$$

Factor x^{2}-\frac{46}{3}x+\frac{529}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{23}{3}\right)^{2}=\frac{361}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{23}{3}\right)^{2}}=\sqrt{\frac{361}{9}}$$

Simplify.

$$x-\frac{23}{3}=\frac{19}{3} x-\frac{23}{3}=-\frac{19}{3}$$

Add \frac{23}{3} to both sides of the equation.

$$x=14 x=\frac{4}{3}$$