Solve [2/3(x-3)]^2=16*7-x/2

$$[ \frac { 2 } { 3 } ( x - 3 ) ] ^ { 2 } = 16 \times \frac { 7 - x } { 2 }$$

Solve for x

$$x=3\sqrt{17}-6\approx 6.369316877$$

$$x=-3\sqrt{17}-6\approx -18.369316877$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

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

Use the distributive property to multiply \frac{2}{3} by x-3.

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

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

$$2\left(\frac{4}{9}x^{2}-\frac{8}{3}x+4\right)=16\left(7-x\right)$$

Use the distributive property to multiply 2 by \frac{4}{9}x^{2}-\frac{8}{3}x+4.

$$\frac{8}{9}x^{2}-\frac{16}{3}x+8=16\left(7-x\right)$$

Use the distributive property to multiply 16 by 7-x.

$$\frac{8}{9}x^{2}-\frac{16}{3}x+8=112-16x$$

Subtract 112 from both sides.

$$\frac{8}{9}x^{2}-\frac{16}{3}x+8-112=-16x$$

Subtract 112 from 8 to get -104.

$$\frac{8}{9}x^{2}-\frac{16}{3}x-104=-16x$$

Add 16x to both sides.

$$\frac{8}{9}x^{2}-\frac{16}{3}x-104+16x=0$$

Combine -\frac{16}{3}x and 16x to get \frac{32}{3}x.

$$\frac{8}{9}x^{2}+\frac{32}{3}x-104=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{8}{9} for a, \frac{32}{3} for b, and -104 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{32}{3}±\sqrt{\left(\frac{32}{3}\right)^{2}-4\times \frac{8}{9}\left(-104\right)}}{2\times \frac{8}{9}}$$

Square \frac{32}{3} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{32}{3}±\sqrt{\frac{1024}{9}-4\times \frac{8}{9}\left(-104\right)}}{2\times \frac{8}{9}}$$

Multiply -4 times \frac{8}{9}.

$$x=\frac{-\frac{32}{3}±\sqrt{\frac{1024}{9}-\frac{32}{9}\left(-104\right)}}{2\times \frac{8}{9}}$$

Multiply -\frac{32}{9} times -104.

$$x=\frac{-\frac{32}{3}±\sqrt{\frac{1024+3328}{9}}}{2\times \frac{8}{9}}$$

Add \frac{1024}{9} to \frac{3328}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{32}{3}±\sqrt{\frac{4352}{9}}}{2\times \frac{8}{9}}$$

Take the square root of \frac{4352}{9}.

$$x=\frac{-\frac{32}{3}±\frac{16\sqrt{17}}{3}}{2\times \frac{8}{9}}$$

Multiply 2 times \frac{8}{9}.

$$x=\frac{-\frac{32}{3}±\frac{16\sqrt{17}}{3}}{\frac{16}{9}}$$

Now solve the equation x=\frac{-\frac{32}{3}±\frac{16\sqrt{17}}{3}}{\frac{16}{9}} when ± is plus. Add -\frac{32}{3} to \frac{16\sqrt{17}}{3}.

$$x=\frac{16\sqrt{17}-32}{\frac{16}{9}\times 3}$$

Divide \frac{-32+16\sqrt{17}}{3} by \frac{16}{9} by multiplying \frac{-32+16\sqrt{17}}{3} by the reciprocal of \frac{16}{9}.

$$x=3\sqrt{17}-6$$

Now solve the equation x=\frac{-\frac{32}{3}±\frac{16\sqrt{17}}{3}}{\frac{16}{9}} when ± is minus. Subtract \frac{16\sqrt{17}}{3} from -\frac{32}{3}.

$$x=\frac{-16\sqrt{17}-32}{\frac{16}{9}\times 3}$$

Divide \frac{-32-16\sqrt{17}}{3} by \frac{16}{9} by multiplying \frac{-32-16\sqrt{17}}{3} by the reciprocal of \frac{16}{9}.

$$x=-3\sqrt{17}-6$$

The equation is now solved.

$$x=3\sqrt{17}-6 x=-3\sqrt{17}-6$$

Steps for Completing the Square

Multiply both sides of the equation by 2.

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

Use the distributive property to multiply \frac{2}{3} by x-3.

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

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

$$2\left(\frac{4}{9}x^{2}-\frac{8}{3}x+4\right)=16\left(7-x\right)$$

Use the distributive property to multiply 2 by \frac{4}{9}x^{2}-\frac{8}{3}x+4.

$$\frac{8}{9}x^{2}-\frac{16}{3}x+8=16\left(7-x\right)$$

Use the distributive property to multiply 16 by 7-x.

$$\frac{8}{9}x^{2}-\frac{16}{3}x+8=112-16x$$

Add 16x to both sides.

$$\frac{8}{9}x^{2}-\frac{16}{3}x+8+16x=112$$

Combine -\frac{16}{3}x and 16x to get \frac{32}{3}x.

$$\frac{8}{9}x^{2}+\frac{32}{3}x+8=112$$

Subtract 8 from both sides.

$$\frac{8}{9}x^{2}+\frac{32}{3}x=112-8$$

Subtract 8 from 112 to get 104.

$$\frac{8}{9}x^{2}+\frac{32}{3}x=104$$

Divide both sides of the equation by \frac{8}{9}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{8}{9}x^{2}+\frac{32}{3}x}{\frac{8}{9}}=\frac{104}{\frac{8}{9}}$$

Dividing by \frac{8}{9} undoes the multiplication by \frac{8}{9}.

$$x^{2}+\frac{\frac{32}{3}}{\frac{8}{9}}x=\frac{104}{\frac{8}{9}}$$

Divide \frac{32}{3} by \frac{8}{9} by multiplying \frac{32}{3} by the reciprocal of \frac{8}{9}.

$$x^{2}+12x=\frac{104}{\frac{8}{9}}$$

Divide 104 by \frac{8}{9} by multiplying 104 by the reciprocal of \frac{8}{9}.

$$x^{2}+12x=117$$

Divide 12, the coefficient of the x term, by 2 to get 6. Then add the square of 6 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+12x+6^{2}=117+6^{2}$$

Square 6.

$$x^{2}+12x+36=117+36$$

Add 117 to 36.

$$x^{2}+12x+36=153$$

Factor x^{2}+12x+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}.

$$\left(x+6\right)^{2}=153$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+6\right)^{2}}=\sqrt{153}$$

Simplify.

$$x+6=3\sqrt{17} x+6=-3\sqrt{17}$$

Subtract 6 from both sides of the equation.

$$x=3\sqrt{17}-6 x=-3\sqrt{17}-6$$