Solve 4-2x+3=4x^2-4x+1

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

Solve for x

$$x=-1$$

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Steps Using Factoring By Grouping

Add 4 and 3 to get 7.

$$7-2x=4x^{2}-4x+1$$

Subtract 4x^{2} from both sides.

$$7-2x-4x^{2}=-4x+1$$

Add 4x to both sides.

$$7-2x-4x^{2}+4x=1$$

Combine -2x and 4x to get 2x.

$$7+2x-4x^{2}=1$$

Subtract 1 from both sides.

$$7+2x-4x^{2}-1=0$$

Subtract 1 from 7 to get 6.

$$6+2x-4x^{2}=0$$

Divide both sides by 2.

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

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

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

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

$$a+b=1 ab=-2\times 3=-6$$

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 -6.

$$-1,6 -2,3$$

Calculate the sum for each pair.

$$-1+6=5 -2+3=1$$

The solution is the pair that gives sum 1.

$$a=3 b=-2$$

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

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

Factor out -x in the first and -1 in the second group.

$$-x\left(2x-3\right)-\left(2x-3\right)$$

Factor out common term 2x-3 by using distributive property.

$$\left(2x-3\right)\left(-x-1\right)$$

To find equation solutions, solve 2x-3=0 and -x-1=0.

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

Steps Using the Quadratic Formula

Add 4 and 3 to get 7.

$$7-2x=4x^{2}-4x+1$$

Subtract 4x^{2} from both sides.

$$7-2x-4x^{2}=-4x+1$$

Add 4x to both sides.

$$7-2x-4x^{2}+4x=1$$

Combine -2x and 4x to get 2x.

$$7+2x-4x^{2}=1$$

Subtract 1 from both sides.

$$7+2x-4x^{2}-1=0$$

Subtract 1 from 7 to get 6.

$$6+2x-4x^{2}=0$$

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.

$$-4x^{2}+2x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 2 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-4\right)\times 6}}{2\left(-4\right)}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\left(-4\right)\times 6}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-2±\sqrt{4+16\times 6}}{2\left(-4\right)}$$

Multiply 16 times 6.

$$x=\frac{-2±\sqrt{4+96}}{2\left(-4\right)}$$

Add 4 to 96.

$$x=\frac{-2±\sqrt{100}}{2\left(-4\right)}$$

Take the square root of 100.

$$x=\frac{-2±10}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-2±10}{-8}$$

Now solve the equation x=\frac{-2±10}{-8} when ± is plus. Add -2 to 10.

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

Divide 8 by -8.

$$x=-1$$

Now solve the equation x=\frac{-2±10}{-8} when ± is minus. Subtract 10 from -2.

$$x=-\frac{12}{-8}$$

Reduce the fraction \frac{-12}{-8} to lowest terms by extracting and canceling out 4.

$$x=\frac{3}{2}$$

The equation is now solved.

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

Steps for Completing the Square

Add 4 and 3 to get 7.

$$7-2x=4x^{2}-4x+1$$

Subtract 4x^{2} from both sides.

$$7-2x-4x^{2}=-4x+1$$

Add 4x to both sides.

$$7-2x-4x^{2}+4x=1$$

Combine -2x and 4x to get 2x.

$$7+2x-4x^{2}=1$$

Subtract 7 from both sides.

$$2x-4x^{2}=1-7$$

Subtract 7 from 1 to get -6.

$$2x-4x^{2}=-6$$

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.

$$-4x^{2}+2x=-6$$

Divide both sides by -4.

$$\frac{-4x^{2}+2x}{-4}=-\frac{6}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\frac{2}{-4}x=-\frac{6}{-4}$$

Reduce the fraction \frac{2}{-4} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{1}{2}x=-\frac{6}{-4}$$

Reduce the fraction \frac{-6}{-4} to lowest terms by extracting and canceling out 2.

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

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{3}{2}+\frac{1}{16}$$

Add \frac{3}{2} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{25}{16}$$

Factor x^{2}-\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$x-\frac{1}{4}=\frac{5}{4} x-\frac{1}{4}=-\frac{5}{4}$$

Add \frac{1}{4} to both sides of the equation.

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