Solve 4x^2=20x-4

$$4 x ^ { 2 } = 20 x - 4$$

Solve for x

$$x = \frac{\sqrt{21} + 5}{2} \approx 4.791287847$$

$$x=\frac{5-\sqrt{21}}{2}\approx 0.208712153$$

Steps Using the Quadratic Formula

Subtract 20x from both sides.

$$4x^{2}-20x=-4$$

Add 4 to both sides.

$$4x^{2}-20x+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -20 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 4\times 4}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-20\right)±\sqrt{400-16\times 4}}{2\times 4}$$

Multiply -16 times 4.

$$x=\frac{-\left(-20\right)±\sqrt{400-64}}{2\times 4}$$

Add 400 to -64.

$$x=\frac{-\left(-20\right)±\sqrt{336}}{2\times 4}$$

Take the square root of 336.

$$x=\frac{-\left(-20\right)±4\sqrt{21}}{2\times 4}$$

The opposite of -20 is 20.

$$x=\frac{20±4\sqrt{21}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{20±4\sqrt{21}}{8}$$

Now solve the equation x=\frac{20±4\sqrt{21}}{8} when ± is plus. Add 20 to 4\sqrt{21}.

$$x=\frac{4\sqrt{21}+20}{8}$$

Divide 20+4\sqrt{21} by 8.

$$x=\frac{\sqrt{21}+5}{2}$$

Now solve the equation x=\frac{20±4\sqrt{21}}{8} when ± is minus. Subtract 4\sqrt{21} from 20.

$$x=\frac{20-4\sqrt{21}}{8}$$

Divide 20-4\sqrt{21} by 8.

$$x=\frac{5-\sqrt{21}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{21}+5}{2} x=\frac{5-\sqrt{21}}{2}$$

Steps for Completing the Square

Subtract 20x from both sides.

$$4x^{2}-20x=-4$$

Divide both sides by 4.

$$\frac{4x^{2}-20x}{4}=-\frac{4}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{20}{4}\right)x=-\frac{4}{4}$$

Divide -20 by 4.

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

Divide -4 by 4.

$$x^{2}-5x=-1$$

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-1+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=-1+\frac{25}{4}$$

Add -1 to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{21}{4}$$

Factor x^{2}-5x+\frac{25}{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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{21}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{21}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{\sqrt{21}}{2} x-\frac{5}{2}=-\frac{\sqrt{21}}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$x=\frac{\sqrt{21}+5}{2} x=\frac{5-\sqrt{21}}{2}$$