Solve 3-3x=4-4x^2

$$3 - 3 x = 4 - 4 x ^ { 2 }$$

Solve for x

$$x=-\frac{1}{4}=-0.25$$

$$x=1$$

Steps Using Factoring By Grouping

Subtract 4 from both sides.

$$3-3x-4=-4x^{2}$$

Subtract 4 from 3 to get -1.

$$-1-3x=-4x^{2}$$

Add 4x^{2} to both sides.

$$-1-3x+4x^{2}=0$$

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

$$4x^{2}-3x-1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=-3 ab=4\left(-1\right)=-4$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -4.

$$1,-4 2,-2$$

Calculate the sum for each pair.

$$1-4=-3 2-2=0$$

The solution is the pair that gives sum -3.

$$a=-4 b=1$$

Rewrite 4x^{2}-3x-1 as \left(4x^{2}-4x\right)+\left(x-1\right).

$$\left(4x^{2}-4x\right)+\left(x-1\right)$$

Factor out 4x in 4x^{2}-4x.

$$4x\left(x-1\right)+x-1$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(4x+1\right)$$

To find equation solutions, solve x-1=0 and 4x+1=0.

$$x=1 x=-\frac{1}{4}$$

Steps Using the Quadratic Formula

Subtract 4 from both sides.

$$3-3x-4=-4x^{2}$$

Subtract 4 from 3 to get -1.

$$-1-3x=-4x^{2}$$

Add 4x^{2} to both sides.

$$-1-3x+4x^{2}=0$$

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.

$$4x^{2}-3x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -3 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 4\left(-1\right)}}{2\times 4}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 4\left(-1\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-3\right)±\sqrt{9-16\left(-1\right)}}{2\times 4}$$

Multiply -16 times -1.

$$x=\frac{-\left(-3\right)±\sqrt{9+16}}{2\times 4}$$

Add 9 to 16.

$$x=\frac{-\left(-3\right)±\sqrt{25}}{2\times 4}$$

Take the square root of 25.

$$x=\frac{-\left(-3\right)±5}{2\times 4}$$

The opposite of -3 is 3.

$$x=\frac{3±5}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{3±5}{8}$$

Now solve the equation x=\frac{3±5}{8} when ± is plus. Add 3 to 5.

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

Divide 8 by 8.

$$x=1$$

Now solve the equation x=\frac{3±5}{8} when ± is minus. Subtract 5 from 3.

$$x=-\frac{2}{8}$$

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

$$x=-\frac{1}{4}$$

The equation is now solved.

$$x=1 x=-\frac{1}{4}$$

Steps for Completing the Square

Add 4x^{2} to both sides.

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

Subtract 3 from both sides.

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

Subtract 3 from 4 to get 1.

$$-3x+4x^{2}=1$$

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.

$$4x^{2}-3x=1$$

Divide both sides by 4.

$$\frac{4x^{2}-3x}{4}=\frac{1}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}-\frac{3}{4}x=\frac{1}{4}$$

Divide -\frac{3}{4}, the coefficient of the x term, by 2 to get -\frac{3}{8}. Then add the square of -\frac{3}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{4}x+\left(-\frac{3}{8}\right)^{2}=\frac{1}{4}+\left(-\frac{3}{8}\right)^{2}$$

Square -\frac{3}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{4}x+\frac{9}{64}=\frac{1}{4}+\frac{9}{64}$$

Add \frac{1}{4} to \frac{9}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{4}x+\frac{9}{64}=\frac{25}{64}$$

Factor x^{2}-\frac{3}{4}x+\frac{9}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{8}\right)^{2}=\frac{25}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{8}\right)^{2}}=\sqrt{\frac{25}{64}}$$

Simplify.

$$x-\frac{3}{8}=\frac{5}{8} x-\frac{3}{8}=-\frac{5}{8}$$

Add \frac{3}{8} to both sides of the equation.

$$x=1 x=-\frac{1}{4}$$