Solve 3xleft(x-5right)=400

$$3x \left( x-5 \right) = 400$$

Solve for x

$$x=\frac{5\sqrt{201}}{6}+\frac{5}{2}\approx 14.314539066$$

$$x=-\frac{5\sqrt{201}}{6}+\frac{5}{2}\approx -9.314539066$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3x by x-5.

$$3x^{2}-15x=400$$

Subtract 400 from both sides.

$$3x^{2}-15x-400=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -15 for b, and -400 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 3\left(-400\right)}}{2\times 3}$$

Square -15.

$$x=\frac{-\left(-15\right)±\sqrt{225-4\times 3\left(-400\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-15\right)±\sqrt{225-12\left(-400\right)}}{2\times 3}$$

Multiply -12 times -400.

$$x=\frac{-\left(-15\right)±\sqrt{225+4800}}{2\times 3}$$

Add 225 to 4800.

$$x=\frac{-\left(-15\right)±\sqrt{5025}}{2\times 3}$$

Take the square root of 5025.

$$x=\frac{-\left(-15\right)±5\sqrt{201}}{2\times 3}$$

The opposite of -15 is 15.

$$x=\frac{15±5\sqrt{201}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{15±5\sqrt{201}}{6}$$

Now solve the equation x=\frac{15±5\sqrt{201}}{6} when ± is plus. Add 15 to 5\sqrt{201}.

$$x=\frac{5\sqrt{201}+15}{6}$$

Divide 15+5\sqrt{201} by 6.

$$x=\frac{5\sqrt{201}}{6}+\frac{5}{2}$$

Now solve the equation x=\frac{15±5\sqrt{201}}{6} when ± is minus. Subtract 5\sqrt{201} from 15.

$$x=\frac{15-5\sqrt{201}}{6}$$

Divide 15-5\sqrt{201} by 6.

$$x=-\frac{5\sqrt{201}}{6}+\frac{5}{2}$$

The equation is now solved.

$$x=\frac{5\sqrt{201}}{6}+\frac{5}{2} x=-\frac{5\sqrt{201}}{6}+\frac{5}{2}$$

Steps for Completing the Square

Use the distributive property to multiply 3x by x-5.

$$3x^{2}-15x=400$$

Divide both sides by 3.

$$\frac{3x^{2}-15x}{3}=\frac{400}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\left(-\frac{15}{3}\right)x=\frac{400}{3}$$

Divide -15 by 3.

$$x^{2}-5x=\frac{400}{3}$$

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.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=\frac{400}{3}+\left(-\frac{5}{2}\right)^{2}$$

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

$$x^{2}-5x+\frac{25}{4}=\frac{400}{3}+\frac{25}{4}$$

Add \frac{400}{3} to \frac{25}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-5x+\frac{25}{4}=\frac{1675}{12}$$

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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{1675}{12}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{1675}{12}}$$

Simplify.

$$x-\frac{5}{2}=\frac{5\sqrt{201}}{6} x-\frac{5}{2}=-\frac{5\sqrt{201}}{6}$$

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

$$x=\frac{5\sqrt{201}}{6}+\frac{5}{2} x=-\frac{5\sqrt{201}}{6}+\frac{5}{2}$$