Solve 0=4x^2-28x+25

$$0=4 { x }^{ 2 } -28x+25$$

Solve for x

$$x=\sqrt{6}+\frac{7}{2}\approx 5.949489743$$

$$x=\frac{7}{2}-\sqrt{6}\approx 1.050510257$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$4x^{2}-28x+25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -28 for b, and 25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}-4\times 4\times 25}}{2\times 4}$$

Square -28.

$$x=\frac{-\left(-28\right)±\sqrt{784-4\times 4\times 25}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-28\right)±\sqrt{784-16\times 25}}{2\times 4}$$

Multiply -16 times 25.

$$x=\frac{-\left(-28\right)±\sqrt{784-400}}{2\times 4}$$

Add 784 to -400.

$$x=\frac{-\left(-28\right)±\sqrt{384}}{2\times 4}$$

Take the square root of 384.

$$x=\frac{-\left(-28\right)±8\sqrt{6}}{2\times 4}$$

The opposite of -28 is 28.

$$x=\frac{28±8\sqrt{6}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{28±8\sqrt{6}}{8}$$

Now solve the equation x=\frac{28±8\sqrt{6}}{8} when ± is plus. Add 28 to 8\sqrt{6}.

$$x=\frac{8\sqrt{6}+28}{8}$$

Divide 28+8\sqrt{6} by 8.

$$x=\sqrt{6}+\frac{7}{2}$$

Now solve the equation x=\frac{28±8\sqrt{6}}{8} when ± is minus. Subtract 8\sqrt{6} from 28.

$$x=\frac{28-8\sqrt{6}}{8}$$

Divide 28-8\sqrt{6} by 8.

$$x=\frac{7}{2}-\sqrt{6}$$

The equation is now solved.

$$x=\sqrt{6}+\frac{7}{2} x=\frac{7}{2}-\sqrt{6}$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$4x^{2}-28x+25=0$$

Subtract 25 from both sides. Anything subtracted from zero gives its negation.

$$4x^{2}-28x=-25$$

Divide both sides by 4.

$$\frac{4x^{2}-28x}{4}=-\frac{25}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{28}{4}\right)x=-\frac{25}{4}$$

Divide -28 by 4.

$$x^{2}-7x=-\frac{25}{4}$$

Divide -7, the coefficient of the x term, by 2 to get -\frac{7}{2}. Then add the square of -\frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-7x+\left(-\frac{7}{2}\right)^{2}=-\frac{25}{4}+\left(-\frac{7}{2}\right)^{2}$$

Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-7x+\frac{49}{4}=\frac{-25+49}{4}$$

Add -\frac{25}{4} to \frac{49}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-7x+\frac{49}{4}=6$$

Factor x^{2}-7x+\frac{49}{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}.

$$\left(x-\frac{7}{2}\right)^{2}=6$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{2}\right)^{2}}=\sqrt{6}$$

Simplify.

$$x-\frac{7}{2}=\sqrt{6} x-\frac{7}{2}=-\sqrt{6}$$

Add \frac{7}{2} to both sides of the equation.

$$x=\sqrt{6}+\frac{7}{2} x=\frac{7}{2}-\sqrt{6}$$