Solve left(5xright)^2-4x-5=0

$${ \left(5x \right) }^{ 2 } -4x-5 = 0$$

Solve for x

$$x=\frac{\sqrt{129}+2}{25}\approx 0.534312668$$

$$x=\frac{2-\sqrt{129}}{25}\approx -0.374312668$$

Steps Using the Quadratic Formula

Expand \left(5x\right)^{2}.

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

Calculate 5 to the power of 2 and get 25.

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

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

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

Square -4.

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

Multiply -4 times 25.

$$x=\frac{-\left(-4\right)±\sqrt{16-100\left(-5\right)}}{2\times 25}$$

Multiply -100 times -5.

$$x=\frac{-\left(-4\right)±\sqrt{16+500}}{2\times 25}$$

Add 16 to 500.

$$x=\frac{-\left(-4\right)±\sqrt{516}}{2\times 25}$$

Take the square root of 516.

$$x=\frac{-\left(-4\right)±2\sqrt{129}}{2\times 25}$$

The opposite of -4 is 4.

$$x=\frac{4±2\sqrt{129}}{2\times 25}$$

Multiply 2 times 25.

$$x=\frac{4±2\sqrt{129}}{50}$$

Now solve the equation x=\frac{4±2\sqrt{129}}{50} when ± is plus. Add 4 to 2\sqrt{129}.

$$x=\frac{2\sqrt{129}+4}{50}$$

Divide 4+2\sqrt{129} by 50.

$$x=\frac{\sqrt{129}+2}{25}$$

Now solve the equation x=\frac{4±2\sqrt{129}}{50} when ± is minus. Subtract 2\sqrt{129} from 4.

$$x=\frac{4-2\sqrt{129}}{50}$$

Divide 4-2\sqrt{129} by 50.

$$x=\frac{2-\sqrt{129}}{25}$$

The equation is now solved.

$$x=\frac{\sqrt{129}+2}{25} x=\frac{2-\sqrt{129}}{25}$$

Steps for Completing the Square

Expand \left(5x\right)^{2}.

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

Calculate 5 to the power of 2 and get 25.

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

Add 5 to both sides. Anything plus zero gives itself.

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

Divide both sides by 25.

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

Dividing by 25 undoes the multiplication by 25.

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

Reduce the fraction \frac{5}{25} to lowest terms by extracting and canceling out 5.

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

Divide -\frac{4}{25}, the coefficient of the x term, by 2 to get -\frac{2}{25}. Then add the square of -\frac{2}{25} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{4}{25}x+\left(-\frac{2}{25}\right)^{2}=\frac{1}{5}+\left(-\frac{2}{25}\right)^{2}$$

Square -\frac{2}{25} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{4}{25}x+\frac{4}{625}=\frac{1}{5}+\frac{4}{625}$$

Add \frac{1}{5} to \frac{4}{625} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{4}{25}x+\frac{4}{625}=\frac{129}{625}$$

Factor x^{2}-\frac{4}{25}x+\frac{4}{625}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{2}{25}\right)^{2}=\frac{129}{625}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{25}\right)^{2}}=\sqrt{\frac{129}{625}}$$

Simplify.

$$x-\frac{2}{25}=\frac{\sqrt{129}}{25} x-\frac{2}{25}=-\frac{\sqrt{129}}{25}$$

Add \frac{2}{25} to both sides of the equation.

$$x=\frac{\sqrt{129}+2}{25} x=\frac{2-\sqrt{129}}{25}$$