Solve 4x^2/5-x-5=0

$$\frac { 4 x ^ { 2 } } { 5 } - x - 5 = 0$$

Solve for x

$$x = \frac{5 \sqrt{17} + 5}{8} \approx 3.201941016$$

$$x=\frac{5-5\sqrt{17}}{8}\approx -1.951941016$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 5.

$$4x^{2}-5x-25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -5 for b, and -25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 4\left(-25\right)}}{2\times 4}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 4\left(-25\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-5\right)±\sqrt{25-16\left(-25\right)}}{2\times 4}$$

Multiply -16 times -25.

$$x=\frac{-\left(-5\right)±\sqrt{25+400}}{2\times 4}$$

Add 25 to 400.

$$x=\frac{-\left(-5\right)±\sqrt{425}}{2\times 4}$$

Take the square root of 425.

$$x=\frac{-\left(-5\right)±5\sqrt{17}}{2\times 4}$$

The opposite of -5 is 5.

$$x=\frac{5±5\sqrt{17}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{5±5\sqrt{17}}{8}$$

Now solve the equation x=\frac{5±5\sqrt{17}}{8} when ± is plus. Add 5 to 5\sqrt{17}.

$$x=\frac{5\sqrt{17}+5}{8}$$

Now solve the equation x=\frac{5±5\sqrt{17}}{8} when ± is minus. Subtract 5\sqrt{17} from 5.

$$x=\frac{5-5\sqrt{17}}{8}$$

The equation is now solved.

$$x=\frac{5\sqrt{17}+5}{8} x=\frac{5-5\sqrt{17}}{8}$$

Steps for Completing the Square

Multiply both sides of the equation by 5.

$$4x^{2}-5x-25=0$$

Add 25 to both sides. Anything plus zero gives itself.

$$4x^{2}-5x=25$$

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

Divide -\frac{5}{4}, the coefficient of the x term, by 2 to get -\frac{5}{8}. Then add the square of -\frac{5}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{4}x+\left(-\frac{5}{8}\right)^{2}=\frac{25}{4}+\left(-\frac{5}{8}\right)^{2}$$

Square -\frac{5}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{4}x+\frac{25}{64}=\frac{25}{4}+\frac{25}{64}$$

Add \frac{25}{4} to \frac{25}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{5}{4}x+\frac{25}{64}=\frac{425}{64}$$

Factor x^{2}-\frac{5}{4}x+\frac{25}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{8}\right)^{2}=\frac{425}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{8}\right)^{2}}=\sqrt{\frac{425}{64}}$$

Simplify.

$$x-\frac{5}{8}=\frac{5\sqrt{17}}{8} x-\frac{5}{8}=-\frac{5\sqrt{17}}{8}$$

Add \frac{5}{8} to both sides of the equation.

$$x=\frac{5\sqrt{17}+5}{8} x=\frac{5-5\sqrt{17}}{8}$$