Solve {left(sqrt{(x+14)*3x}right)}^2=x+01

$${ \left( \sqrt{ (x+14) \times 3x } \right) }^{ 2 } =x+01$$

Solve for x (complex solution)

$$x = -\frac{41}{3} = -13\frac{2}{3} \approx -13.666666667$$

$$x=0$$

Steps Using Factoring

Use the distributive property to multiply x+14 by 3.

$$\left(\sqrt{\left(3x+42\right)x}\right)^{2}=x+0\times 1$$

Use the distributive property to multiply 3x+42 by x.

$$\left(\sqrt{3x^{2}+42x}\right)^{2}=x+0\times 1$$

Calculate \sqrt{3x^{2}+42x} to the power of 2 and get 3x^{2}+42x.

$$3x^{2}+42x=x+0\times 1$$

Multiply 0 and 1 to get 0.

$$3x^{2}+42x=x+0$$

Anything plus zero gives itself.

$$3x^{2}+42x=x$$

Subtract x from both sides.

$$3x^{2}+42x-x=0$$

Combine 42x and -x to get 41x.

$$3x^{2}+41x=0$$

Factor out x.

$$x\left(3x+41\right)=0$$

To find equation solutions, solve x=0 and 3x+41=0.

$$x=0 x=-\frac{41}{3}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x+14 by 3.

$$\left(\sqrt{\left(3x+42\right)x}\right)^{2}=x+0\times 1$$

Use the distributive property to multiply 3x+42 by x.

$$\left(\sqrt{3x^{2}+42x}\right)^{2}=x+0\times 1$$

Calculate \sqrt{3x^{2}+42x} to the power of 2 and get 3x^{2}+42x.

$$3x^{2}+42x=x+0\times 1$$

Multiply 0 and 1 to get 0.

$$3x^{2}+42x=x+0$$

Anything plus zero gives itself.

$$3x^{2}+42x=x$$

Subtract x from both sides.

$$3x^{2}+42x-x=0$$

Combine 42x and -x to get 41x.

$$3x^{2}+41x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 41 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-41±\sqrt{41^{2}}}{2\times 3}$$

Take the square root of 41^{2}.

$$x=\frac{-41±41}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-41±41}{6}$$

Now solve the equation x=\frac{-41±41}{6} when ± is plus. Add -41 to 41.

$$x=\frac{0}{6}$$

Divide 0 by 6.

$$x=0$$

Now solve the equation x=\frac{-41±41}{6} when ± is minus. Subtract 41 from -41.

$$x=-\frac{82}{6}$$

Reduce the fraction \frac{-82}{6} to lowest terms by extracting and canceling out 2.

$$x=-\frac{41}{3}$$

The equation is now solved.

$$x=0 x=-\frac{41}{3}$$

Steps for Completing the Square

Use the distributive property to multiply x+14 by 3.

$$\left(\sqrt{\left(3x+42\right)x}\right)^{2}=x+0\times 1$$

Use the distributive property to multiply 3x+42 by x.

$$\left(\sqrt{3x^{2}+42x}\right)^{2}=x+0\times 1$$

Calculate \sqrt{3x^{2}+42x} to the power of 2 and get 3x^{2}+42x.

$$3x^{2}+42x=x+0\times 1$$

Multiply 0 and 1 to get 0.

$$3x^{2}+42x=x+0$$

Anything plus zero gives itself.

$$3x^{2}+42x=x$$

Subtract x from both sides.

$$3x^{2}+42x-x=0$$

Combine 42x and -x to get 41x.

$$3x^{2}+41x=0$$

Divide both sides by 3.

$$\frac{3x^{2}+41x}{3}=\frac{0}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{41}{3}x=\frac{0}{3}$$

Divide 0 by 3.

$$x^{2}+\frac{41}{3}x=0$$

Divide \frac{41}{3}, the coefficient of the x term, by 2 to get \frac{41}{6}. Then add the square of \frac{41}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{41}{3}x+\left(\frac{41}{6}\right)^{2}=\left(\frac{41}{6}\right)^{2}$$

Square \frac{41}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{41}{3}x+\frac{1681}{36}=\frac{1681}{36}$$

Factor x^{2}+\frac{41}{3}x+\frac{1681}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{41}{6}\right)^{2}=\frac{1681}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{41}{6}\right)^{2}}=\sqrt{\frac{1681}{36}}$$

Simplify.

$$x+\frac{41}{6}=\frac{41}{6} x+\frac{41}{6}=-\frac{41}{6}$$

Subtract \frac{41}{6} from both sides of the equation.

$$x=0 x=-\frac{41}{3}$$

Solve for x

$$x=0$$

Steps Using Factoring

Use the distributive property to multiply x+14 by 3.

$$\left(\sqrt{\left(3x+42\right)x}\right)^{2}=x+0\times 1$$

Use the distributive property to multiply 3x+42 by x.

$$\left(\sqrt{3x^{2}+42x}\right)^{2}=x+0\times 1$$

Calculate \sqrt{3x^{2}+42x} to the power of 2 and get 3x^{2}+42x.

$$3x^{2}+42x=x+0\times 1$$

Multiply 0 and 1 to get 0.

$$3x^{2}+42x=x+0$$

Anything plus zero gives itself.

$$3x^{2}+42x=x$$

Subtract x from both sides.

$$3x^{2}+42x-x=0$$

Combine 42x and -x to get 41x.

$$3x^{2}+41x=0$$

Factor out x.

$$x\left(3x+41\right)=0$$

To find equation solutions, solve x=0 and 3x+41=0.

$$x=0 x=-\frac{41}{3}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x+14 by 3.

$$\left(\sqrt{\left(3x+42\right)x}\right)^{2}=x+0\times 1$$

Use the distributive property to multiply 3x+42 by x.

$$\left(\sqrt{3x^{2}+42x}\right)^{2}=x+0\times 1$$

Calculate \sqrt{3x^{2}+42x} to the power of 2 and get 3x^{2}+42x.

$$3x^{2}+42x=x+0\times 1$$

Multiply 0 and 1 to get 0.

$$3x^{2}+42x=x+0$$

Anything plus zero gives itself.

$$3x^{2}+42x=x$$

Subtract x from both sides.

$$3x^{2}+42x-x=0$$

Combine 42x and -x to get 41x.

$$3x^{2}+41x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 41 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-41±\sqrt{41^{2}}}{2\times 3}$$

Take the square root of 41^{2}.

$$x=\frac{-41±41}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-41±41}{6}$$

Now solve the equation x=\frac{-41±41}{6} when ± is plus. Add -41 to 41.

$$x=\frac{0}{6}$$

Divide 0 by 6.

$$x=0$$

Now solve the equation x=\frac{-41±41}{6} when ± is minus. Subtract 41 from -41.

$$x=-\frac{82}{6}$$

Reduce the fraction \frac{-82}{6} to lowest terms by extracting and canceling out 2.

$$x=-\frac{41}{3}$$

The equation is now solved.

$$x=0 x=-\frac{41}{3}$$

Steps for Completing the Square

Use the distributive property to multiply x+14 by 3.

$$\left(\sqrt{\left(3x+42\right)x}\right)^{2}=x+0\times 1$$

Use the distributive property to multiply 3x+42 by x.

$$\left(\sqrt{3x^{2}+42x}\right)^{2}=x+0\times 1$$

Calculate \sqrt{3x^{2}+42x} to the power of 2 and get 3x^{2}+42x.

$$3x^{2}+42x=x+0\times 1$$

Multiply 0 and 1 to get 0.

$$3x^{2}+42x=x+0$$

Anything plus zero gives itself.

$$3x^{2}+42x=x$$

Subtract x from both sides.

$$3x^{2}+42x-x=0$$

Combine 42x and -x to get 41x.

$$3x^{2}+41x=0$$

Divide both sides by 3.

$$\frac{3x^{2}+41x}{3}=\frac{0}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{41}{3}x=\frac{0}{3}$$

Divide 0 by 3.

$$x^{2}+\frac{41}{3}x=0$$

Divide \frac{41}{3}, the coefficient of the x term, by 2 to get \frac{41}{6}. Then add the square of \frac{41}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{41}{3}x+\left(\frac{41}{6}\right)^{2}=\left(\frac{41}{6}\right)^{2}$$

Square \frac{41}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{41}{3}x+\frac{1681}{36}=\frac{1681}{36}$$

Factor x^{2}+\frac{41}{3}x+\frac{1681}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{41}{6}\right)^{2}=\frac{1681}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{41}{6}\right)^{2}}=\sqrt{\frac{1681}{36}}$$

Simplify.

$$x+\frac{41}{6}=\frac{41}{6} x+\frac{41}{6}=-\frac{41}{6}$$

Subtract \frac{41}{6} from both sides of the equation.

$$x=0 x=-\frac{41}{3}$$