Solve 8x+12=4x^2-4x+1

$$8x+12 = 4 { x }^{ 2 } -4x+1$$

Solve for x

$$x=\sqrt{5}+\frac{3}{2}\approx 3.736067977$$

$$x=\frac{3}{2}-\sqrt{5}\approx -0.736067977$$

Steps Using the Quadratic Formula

Subtract 4x^{2} from both sides.

$$8x+12-4x^{2}=-4x+1$$

Add 4x to both sides.

$$8x+12-4x^{2}+4x=1$$

Combine 8x and 4x to get 12x.

$$12x+12-4x^{2}=1$$

Subtract 1 from both sides.

$$12x+12-4x^{2}-1=0$$

Subtract 1 from 12 to get 11.

$$12x+11-4x^{2}=0$$

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.

$$-4x^{2}+12x+11=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 12 for b, and 11 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\left(-4\right)\times 11}}{2\left(-4\right)}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\left(-4\right)\times 11}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-12±\sqrt{144+16\times 11}}{2\left(-4\right)}$$

Multiply 16 times 11.

$$x=\frac{-12±\sqrt{144+176}}{2\left(-4\right)}$$

Add 144 to 176.

$$x=\frac{-12±\sqrt{320}}{2\left(-4\right)}$$

Take the square root of 320.

$$x=\frac{-12±8\sqrt{5}}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-12±8\sqrt{5}}{-8}$$

Now solve the equation x=\frac{-12±8\sqrt{5}}{-8} when ± is plus. Add -12 to 8\sqrt{5}.

$$x=\frac{8\sqrt{5}-12}{-8}$$

Divide -12+8\sqrt{5} by -8.

$$x=\frac{3}{2}-\sqrt{5}$$

Now solve the equation x=\frac{-12±8\sqrt{5}}{-8} when ± is minus. Subtract 8\sqrt{5} from -12.

$$x=\frac{-8\sqrt{5}-12}{-8}$$

Divide -12-8\sqrt{5} by -8.

$$x=\sqrt{5}+\frac{3}{2}$$

The equation is now solved.

$$x=\frac{3}{2}-\sqrt{5} x=\sqrt{5}+\frac{3}{2}$$

Steps for Completing the Square

Subtract 4x^{2} from both sides.

$$8x+12-4x^{2}=-4x+1$$

Add 4x to both sides.

$$8x+12-4x^{2}+4x=1$$

Combine 8x and 4x to get 12x.

$$12x+12-4x^{2}=1$$

Subtract 12 from both sides.

$$12x-4x^{2}=1-12$$

Subtract 12 from 1 to get -11.

$$12x-4x^{2}=-11$$

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.

$$-4x^{2}+12x=-11$$

Divide both sides by -4.

$$\frac{-4x^{2}+12x}{-4}=-\frac{11}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\frac{12}{-4}x=-\frac{11}{-4}$$

Divide 12 by -4.

$$x^{2}-3x=-\frac{11}{-4}$$

Divide -11 by -4.

$$x^{2}-3x=\frac{11}{4}$$

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

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=\frac{11}{4}+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=\frac{11+9}{4}$$

Add \frac{11}{4} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-3x+\frac{9}{4}=5$$

Factor x^{2}-3x+\frac{9}{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}.

$$\left(x-\frac{3}{2}\right)^{2}=5$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{5}$$

Simplify.

$$x-\frac{3}{2}=\sqrt{5} x-\frac{3}{2}=-\sqrt{5}$$

Add \frac{3}{2} to both sides of the equation.

$$x=\sqrt{5}+\frac{3}{2} x=\frac{3}{2}-\sqrt{5}$$