Solve 12x^2+5=16x

$$12 x ^ { 2 } + 5 = 16 x$$

Solve for x

$$x=\frac{1}{2}=0.5$$

$$x=\frac{5}{6}\approx 0.833333333$$

Steps Using Factoring By Grouping

Subtract 16x from both sides.

$$12x^{2}+5-16x=0$$

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

$$12x^{2}-16x+5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 12x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=-16 ab=12\times 5=60$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 60.

$$-1,-60 -2,-30 -3,-20 -4,-15 -5,-12 -6,-10$$

Calculate the sum for each pair.

$$-1-60=-61 -2-30=-32 -3-20=-23 -4-15=-19 -5-12=-17 -6-10=-16$$

The solution is the pair that gives sum -16.

$$a=-10 b=-6$$

Rewrite 12x^{2}-16x+5 as \left(12x^{2}-10x\right)+\left(-6x+5\right).

$$\left(12x^{2}-10x\right)+\left(-6x+5\right)$$

Factor out 2x in the first and -1 in the second group.

$$2x\left(6x-5\right)-\left(6x-5\right)$$

Factor out common term 6x-5 by using distributive property.

$$\left(6x-5\right)\left(2x-1\right)$$

To find equation solutions, solve 6x-5=0 and 2x-1=0.

$$x=\frac{5}{6} x=\frac{1}{2}$$

Steps Using the Quadratic Formula

Subtract 16x from both sides.

$$12x^{2}+5-16x=0$$

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.

$$12x^{2}-16x+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 12 for a, -16 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 12\times 5}}{2\times 12}$$

Square -16.

$$x=\frac{-\left(-16\right)±\sqrt{256-4\times 12\times 5}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-\left(-16\right)±\sqrt{256-48\times 5}}{2\times 12}$$

Multiply -48 times 5.

$$x=\frac{-\left(-16\right)±\sqrt{256-240}}{2\times 12}$$

Add 256 to -240.

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

Take the square root of 16.

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

The opposite of -16 is 16.

$$x=\frac{16±4}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{16±4}{24}$$

Now solve the equation x=\frac{16±4}{24} when ± is plus. Add 16 to 4.

$$x=\frac{20}{24}$$

Reduce the fraction \frac{20}{24} to lowest terms by extracting and canceling out 4.

$$x=\frac{5}{6}$$

Now solve the equation x=\frac{16±4}{24} when ± is minus. Subtract 4 from 16.

$$x=\frac{12}{24}$$

Reduce the fraction \frac{12}{24} to lowest terms by extracting and canceling out 12.

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

The equation is now solved.

$$x=\frac{5}{6} x=\frac{1}{2}$$

Steps for Completing the Square

Subtract 16x from both sides.

$$12x^{2}+5-16x=0$$

Subtract 5 from both sides. Anything subtracted from zero gives its negation.

$$12x^{2}-16x=-5$$

Divide both sides by 12.

$$\frac{12x^{2}-16x}{12}=-\frac{5}{12}$$

Dividing by 12 undoes the multiplication by 12.

$$x^{2}+\left(-\frac{16}{12}\right)x=-\frac{5}{12}$$

Reduce the fraction \frac{-16}{12} to lowest terms by extracting and canceling out 4.

$$x^{2}-\frac{4}{3}x=-\frac{5}{12}$$

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

$$x^{2}-\frac{4}{3}x+\left(-\frac{2}{3}\right)^{2}=-\frac{5}{12}+\left(-\frac{2}{3}\right)^{2}$$

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

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=-\frac{5}{12}+\frac{4}{9}$$

Add -\frac{5}{12} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=\frac{1}{36}$$

Factor x^{2}-\frac{4}{3}x+\frac{4}{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}.

$$\left(x-\frac{2}{3}\right)^{2}=\frac{1}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{3}\right)^{2}}=\sqrt{\frac{1}{36}}$$

Simplify.

$$x-\frac{2}{3}=\frac{1}{6} x-\frac{2}{3}=-\frac{1}{6}$$

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

$$x=\frac{5}{6} x=\frac{1}{2}$$