Solve left(5xright)^2+5x+96=0

$${ \left(5x \right) }^{ 2 } +5x+96 = 0$$

Solve for x (complex solution)

$$x=\frac{-1+\sqrt{383}i}{10}\approx -0.1+1.957038579i$$

$$x=\frac{-\sqrt{383}i-1}{10}\approx -0.1-1.957038579i$$

Steps Using the Quadratic Formula

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

$$5^{2}x^{2}+5x+96=0$$

Calculate 5 to the power of 2 and get 25.

$$25x^{2}+5x+96=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 25 for a, 5 for b, and 96 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\times 25\times 96}}{2\times 25}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\times 25\times 96}}{2\times 25}$$

Multiply -4 times 25.

$$x=\frac{-5±\sqrt{25-100\times 96}}{2\times 25}$$

Multiply -100 times 96.

$$x=\frac{-5±\sqrt{25-9600}}{2\times 25}$$

Add 25 to -9600.

$$x=\frac{-5±\sqrt{-9575}}{2\times 25}$$

Take the square root of -9575.

$$x=\frac{-5±5\sqrt{383}i}{2\times 25}$$

Multiply 2 times 25.

$$x=\frac{-5±5\sqrt{383}i}{50}$$

Now solve the equation x=\frac{-5±5\sqrt{383}i}{50} when ± is plus. Add -5 to 5i\sqrt{383}.

$$x=\frac{-5+5\sqrt{383}i}{50}$$

Divide -5+5i\sqrt{383} by 50.

$$x=\frac{-1+\sqrt{383}i}{10}$$

Now solve the equation x=\frac{-5±5\sqrt{383}i}{50} when ± is minus. Subtract 5i\sqrt{383} from -5.

$$x=\frac{-5\sqrt{383}i-5}{50}$$

Divide -5-5i\sqrt{383} by 50.

$$x=\frac{-\sqrt{383}i-1}{10}$$

The equation is now solved.

$$x=\frac{-1+\sqrt{383}i}{10} x=\frac{-\sqrt{383}i-1}{10}$$

Steps for Completing the Square

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

$$5^{2}x^{2}+5x+96=0$$

Calculate 5 to the power of 2 and get 25.

$$25x^{2}+5x+96=0$$

Subtract 96 from both sides. Anything subtracted from zero gives its negation.

$$25x^{2}+5x=-96$$

Divide both sides by 25.

$$\frac{25x^{2}+5x}{25}=-\frac{96}{25}$$

Dividing by 25 undoes the multiplication by 25.

$$x^{2}+\frac{5}{25}x=-\frac{96}{25}$$

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

$$x^{2}+\frac{1}{5}x=-\frac{96}{25}$$

Divide \frac{1}{5}, the coefficient of the x term, by 2 to get \frac{1}{10}. Then add the square of \frac{1}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{5}x+\left(\frac{1}{10}\right)^{2}=-\frac{96}{25}+\left(\frac{1}{10}\right)^{2}$$

Square \frac{1}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{5}x+\frac{1}{100}=-\frac{96}{25}+\frac{1}{100}$$

Add -\frac{96}{25} to \frac{1}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{5}x+\frac{1}{100}=-\frac{383}{100}$$

Factor x^{2}+\frac{1}{5}x+\frac{1}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{10}\right)^{2}=-\frac{383}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{10}\right)^{2}}=\sqrt{-\frac{383}{100}}$$

Simplify.

$$x+\frac{1}{10}=\frac{\sqrt{383}i}{10} x+\frac{1}{10}=-\frac{\sqrt{383}i}{10}$$

Subtract \frac{1}{10} from both sides of the equation.

$$x=\frac{-1+\sqrt{383}i}{10} x=\frac{-\sqrt{383}i-1}{10}$$