Solve 4x^2+1=6x

$$4 x ^ { 2 } + 1 = 6 x$$

Solve for x

$$x = \frac{\sqrt{5} + 3}{4} \approx 1.309016994$$

$$x=\frac{3-\sqrt{5}}{4}\approx 0.190983006$$

Steps Using the Quadratic Formula

Subtract 6x from both sides.

$$4x^{2}+1-6x=0$$

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.

$$4x^{2}-6x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -6 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 4}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-6\right)±\sqrt{36-16}}{2\times 4}$$

Add 36 to -16.

$$x=\frac{-\left(-6\right)±\sqrt{20}}{2\times 4}$$

Take the square root of 20.

$$x=\frac{-\left(-6\right)±2\sqrt{5}}{2\times 4}$$

The opposite of -6 is 6.

$$x=\frac{6±2\sqrt{5}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{6±2\sqrt{5}}{8}$$

Now solve the equation x=\frac{6±2\sqrt{5}}{8} when ± is plus. Add 6 to 2\sqrt{5}.

$$x=\frac{2\sqrt{5}+6}{8}$$

Divide 6+2\sqrt{5} by 8.

$$x=\frac{\sqrt{5}+3}{4}$$

Now solve the equation x=\frac{6±2\sqrt{5}}{8} when ± is minus. Subtract 2\sqrt{5} from 6.

$$x=\frac{6-2\sqrt{5}}{8}$$

Divide 6-2\sqrt{5} by 8.

$$x=\frac{3-\sqrt{5}}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{5}+3}{4} x=\frac{3-\sqrt{5}}{4}$$

Steps for Completing the Square

Subtract 6x from both sides.

$$4x^{2}+1-6x=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{6}{4}\right)x=-\frac{1}{4}$$

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

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

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

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

Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=-\frac{1}{4}+\frac{9}{16}$$

Add -\frac{1}{4} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{5}{16}$$

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

$$\left(x-\frac{3}{4}\right)^{2}=\frac{5}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{\frac{5}{16}}$$

Simplify.

$$x-\frac{3}{4}=\frac{\sqrt{5}}{4} x-\frac{3}{4}=-\frac{\sqrt{5}}{4}$$

Add \frac{3}{4} to both sides of the equation.

$$x=\frac{\sqrt{5}+3}{4} x=\frac{3-\sqrt{5}}{4}$$