Solve (3x+5/7)^2=16x

$$( \frac { 3 x + 5 } { 7 } ) ^ { 2 } = 16 x$$

Solve for x

$$x = \frac{28 \sqrt{181} + 377}{9} \approx 83.744608146$$

$$x=\frac{377-28\sqrt{181}}{9}\approx 0.033169631$$

Steps Using the Quadratic Formula

To raise \frac{3x+5}{7} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(3x+5\right)^{2}}{7^{2}}=16x$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+5\right)^{2}.

$$\frac{9x^{2}+30x+25}{7^{2}}=16x$$

Calculate 7 to the power of 2 and get 49.

$$\frac{9x^{2}+30x+25}{49}=16x$$

Divide each term of 9x^{2}+30x+25 by 49 to get \frac{9}{49}x^{2}+\frac{30}{49}x+\frac{25}{49}.

$$\frac{9}{49}x^{2}+\frac{30}{49}x+\frac{25}{49}=16x$$

Subtract 16x from both sides.

$$\frac{9}{49}x^{2}+\frac{30}{49}x+\frac{25}{49}-16x=0$$

Combine \frac{30}{49}x and -16x to get -\frac{754}{49}x.

$$\frac{9}{49}x^{2}-\frac{754}{49}x+\frac{25}{49}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{9}{49} for a, -\frac{754}{49} for b, and \frac{25}{49} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{754}{49}\right)±\sqrt{\left(-\frac{754}{49}\right)^{2}-4\times \frac{9}{49}\times \frac{25}{49}}}{2\times \frac{9}{49}}$$

Square -\frac{754}{49} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{754}{49}\right)±\sqrt{\frac{568516}{2401}-4\times \frac{9}{49}\times \frac{25}{49}}}{2\times \frac{9}{49}}$$

Multiply -4 times \frac{9}{49}.

$$x=\frac{-\left(-\frac{754}{49}\right)±\sqrt{\frac{568516}{2401}-\frac{36}{49}\times \frac{25}{49}}}{2\times \frac{9}{49}}$$

Multiply -\frac{36}{49} times \frac{25}{49} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{754}{49}\right)±\sqrt{\frac{568516-900}{2401}}}{2\times \frac{9}{49}}$$

Add \frac{568516}{2401} to -\frac{900}{2401} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{754}{49}\right)±\sqrt{\frac{11584}{49}}}{2\times \frac{9}{49}}$$

Take the square root of \frac{11584}{49}.

$$x=\frac{-\left(-\frac{754}{49}\right)±\frac{8\sqrt{181}}{7}}{2\times \frac{9}{49}}$$

The opposite of -\frac{754}{49} is \frac{754}{49}.

$$x=\frac{\frac{754}{49}±\frac{8\sqrt{181}}{7}}{2\times \frac{9}{49}}$$

Multiply 2 times \frac{9}{49}.

$$x=\frac{\frac{754}{49}±\frac{8\sqrt{181}}{7}}{\frac{18}{49}}$$

Now solve the equation x=\frac{\frac{754}{49}±\frac{8\sqrt{181}}{7}}{\frac{18}{49}} when ± is plus. Add \frac{754}{49} to \frac{8\sqrt{181}}{7}.

$$x=\frac{\frac{8\sqrt{181}}{7}+\frac{754}{49}}{\frac{18}{49}}$$

Divide \frac{754}{49}+\frac{8\sqrt{181}}{7} by \frac{18}{49} by multiplying \frac{754}{49}+\frac{8\sqrt{181}}{7} by the reciprocal of \frac{18}{49}.

$$x=\frac{28\sqrt{181}+377}{9}$$

Now solve the equation x=\frac{\frac{754}{49}±\frac{8\sqrt{181}}{7}}{\frac{18}{49}} when ± is minus. Subtract \frac{8\sqrt{181}}{7} from \frac{754}{49}.

$$x=\frac{-\frac{8\sqrt{181}}{7}+\frac{754}{49}}{\frac{18}{49}}$$

Divide \frac{754}{49}-\frac{8\sqrt{181}}{7} by \frac{18}{49} by multiplying \frac{754}{49}-\frac{8\sqrt{181}}{7} by the reciprocal of \frac{18}{49}.

$$x=\frac{377-28\sqrt{181}}{9}$$

The equation is now solved.

$$x=\frac{28\sqrt{181}+377}{9} x=\frac{377-28\sqrt{181}}{9}$$

Steps for Completing the Square

To raise \frac{3x+5}{7} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(3x+5\right)^{2}}{7^{2}}=16x$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+5\right)^{2}.

$$\frac{9x^{2}+30x+25}{7^{2}}=16x$$

Calculate 7 to the power of 2 and get 49.

$$\frac{9x^{2}+30x+25}{49}=16x$$

Divide each term of 9x^{2}+30x+25 by 49 to get \frac{9}{49}x^{2}+\frac{30}{49}x+\frac{25}{49}.

$$\frac{9}{49}x^{2}+\frac{30}{49}x+\frac{25}{49}=16x$$

Subtract 16x from both sides.

$$\frac{9}{49}x^{2}+\frac{30}{49}x+\frac{25}{49}-16x=0$$

Combine \frac{30}{49}x and -16x to get -\frac{754}{49}x.

$$\frac{9}{49}x^{2}-\frac{754}{49}x+\frac{25}{49}=0$$

Subtract \frac{25}{49} from both sides. Anything subtracted from zero gives its negation.

$$\frac{9}{49}x^{2}-\frac{754}{49}x=-\frac{25}{49}$$

Divide both sides of the equation by \frac{9}{49}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{9}{49}x^{2}-\frac{754}{49}x}{\frac{9}{49}}=-\frac{\frac{25}{49}}{\frac{9}{49}}$$

Dividing by \frac{9}{49} undoes the multiplication by \frac{9}{49}.

$$x^{2}+\left(-\frac{\frac{754}{49}}{\frac{9}{49}}\right)x=-\frac{\frac{25}{49}}{\frac{9}{49}}$$

Divide -\frac{754}{49} by \frac{9}{49} by multiplying -\frac{754}{49} by the reciprocal of \frac{9}{49}.

$$x^{2}-\frac{754}{9}x=-\frac{\frac{25}{49}}{\frac{9}{49}}$$

Divide -\frac{25}{49} by \frac{9}{49} by multiplying -\frac{25}{49} by the reciprocal of \frac{9}{49}.

$$x^{2}-\frac{754}{9}x=-\frac{25}{9}$$

Divide -\frac{754}{9}, the coefficient of the x term, by 2 to get -\frac{377}{9}. Then add the square of -\frac{377}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{754}{9}x+\left(-\frac{377}{9}\right)^{2}=-\frac{25}{9}+\left(-\frac{377}{9}\right)^{2}$$

Square -\frac{377}{9} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{754}{9}x+\frac{142129}{81}=-\frac{25}{9}+\frac{142129}{81}$$

Add -\frac{25}{9} to \frac{142129}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{754}{9}x+\frac{142129}{81}=\frac{141904}{81}$$

Factor x^{2}-\frac{754}{9}x+\frac{142129}{81}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{377}{9}\right)^{2}=\frac{141904}{81}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{377}{9}\right)^{2}}=\sqrt{\frac{141904}{81}}$$

Simplify.

$$x-\frac{377}{9}=\frac{28\sqrt{181}}{9} x-\frac{377}{9}=-\frac{28\sqrt{181}}{9}$$

Add \frac{377}{9} to both sides of the equation.

$$x=\frac{28\sqrt{181}+377}{9} x=\frac{377-28\sqrt{181}}{9}$$