Solve 2(10-3x)(10-3x)-3=29

$$2(10-3x)(10-3x)-3=29$$

Solve for x

$$x=2$$

$$x = \frac{14}{3} = 4\frac{2}{3} \approx 4.666666667$$

Steps Using the Quadratic Formula

Multiply 10-3x and 10-3x to get \left(10-3x\right)^{2}.

$$2\left(10-3x\right)^{2}-3=29$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(10-3x\right)^{2}.

$$2\left(100-60x+9x^{2}\right)-3=29$$

Use the distributive property to multiply 2 by 100-60x+9x^{2}.

$$200-120x+18x^{2}-3=29$$

Subtract 3 from 200 to get 197.

$$197-120x+18x^{2}=29$$

Subtract 29 from both sides.

$$197-120x+18x^{2}-29=0$$

Subtract 29 from 197 to get 168.

$$168-120x+18x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$18x^{2}-120x+168=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 18 for a, -120 for b, and 168 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-120\right)±\sqrt{\left(-120\right)^{2}-4\times 18\times 168}}{2\times 18}$$

Square -120.

$$x=\frac{-\left(-120\right)±\sqrt{14400-4\times 18\times 168}}{2\times 18}$$

Multiply -4 times 18.

$$x=\frac{-\left(-120\right)±\sqrt{14400-72\times 168}}{2\times 18}$$

Multiply -72 times 168.

$$x=\frac{-\left(-120\right)±\sqrt{14400-12096}}{2\times 18}$$

Add 14400 to -12096.

$$x=\frac{-\left(-120\right)±\sqrt{2304}}{2\times 18}$$

Take the square root of 2304.

$$x=\frac{-\left(-120\right)±48}{2\times 18}$$

The opposite of -120 is 120.

$$x=\frac{120±48}{2\times 18}$$

Multiply 2 times 18.

$$x=\frac{120±48}{36}$$

Now solve the equation x=\frac{120±48}{36} when ± is plus. Add 120 to 48.

$$x=\frac{168}{36}$$

Reduce the fraction \frac{168}{36} to lowest terms by extracting and canceling out 12.

$$x=\frac{14}{3}$$

Now solve the equation x=\frac{120±48}{36} when ± is minus. Subtract 48 from 120.

$$x=\frac{72}{36}$$

Divide 72 by 36.

$$x=2$$

The equation is now solved.

$$x=\frac{14}{3} x=2$$

Steps for Completing the Square

Multiply 10-3x and 10-3x to get \left(10-3x\right)^{2}.

$$2\left(10-3x\right)^{2}-3=29$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(10-3x\right)^{2}.

$$2\left(100-60x+9x^{2}\right)-3=29$$

Use the distributive property to multiply 2 by 100-60x+9x^{2}.

$$200-120x+18x^{2}-3=29$$

Subtract 3 from 200 to get 197.

$$197-120x+18x^{2}=29$$

Subtract 197 from both sides.

$$-120x+18x^{2}=29-197$$

Subtract 197 from 29 to get -168.

$$-120x+18x^{2}=-168$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$18x^{2}-120x=-168$$

Divide both sides by 18.

$$\frac{18x^{2}-120x}{18}=-\frac{168}{18}$$

Dividing by 18 undoes the multiplication by 18.

$$x^{2}+\left(-\frac{120}{18}\right)x=-\frac{168}{18}$$

Reduce the fraction \frac{-120}{18} to lowest terms by extracting and canceling out 6.

$$x^{2}-\frac{20}{3}x=-\frac{168}{18}$$

Reduce the fraction \frac{-168}{18} to lowest terms by extracting and canceling out 6.

$$x^{2}-\frac{20}{3}x=-\frac{28}{3}$$

Divide -\frac{20}{3}, the coefficient of the x term, by 2 to get -\frac{10}{3}. Then add the square of -\frac{10}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{20}{3}x+\left(-\frac{10}{3}\right)^{2}=-\frac{28}{3}+\left(-\frac{10}{3}\right)^{2}$$

Square -\frac{10}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{20}{3}x+\frac{100}{9}=-\frac{28}{3}+\frac{100}{9}$$

Add -\frac{28}{3} to \frac{100}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{20}{3}x+\frac{100}{9}=\frac{16}{9}$$

Factor x^{2}-\frac{20}{3}x+\frac{100}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{10}{3}\right)^{2}=\frac{16}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{10}{3}\right)^{2}}=\sqrt{\frac{16}{9}}$$

Simplify.

$$x-\frac{10}{3}=\frac{4}{3} x-\frac{10}{3}=-\frac{4}{3}$$

Add \frac{10}{3} to both sides of the equation.

$$x=\frac{14}{3} x=2$$