Solve 3(7-x)^2-3(x-2)=4(x-3)

$$3 ( 7 - x ) ^ { 2 } - 3 ( x - 2 ) = 4 ( x - 3 )$$

Solve for x

$$x = \frac{\sqrt{421} + 49}{6} \approx 11.586380755$$

$$x = \frac{49 - \sqrt{421}}{6} \approx 4.746952579$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(7-x\right)^{2}.

$$3\left(49-14x+x^{2}\right)-3\left(x-2\right)=4\left(x-3\right)$$

Use the distributive property to multiply 3 by 49-14x+x^{2}.

$$147-42x+3x^{2}-3\left(x-2\right)=4\left(x-3\right)$$

Use the distributive property to multiply -3 by x-2.

$$147-42x+3x^{2}-3x+6=4\left(x-3\right)$$

Combine -42x and -3x to get -45x.

$$147-45x+3x^{2}+6=4\left(x-3\right)$$

Add 147 and 6 to get 153.

$$153-45x+3x^{2}=4\left(x-3\right)$$

Use the distributive property to multiply 4 by x-3.

$$153-45x+3x^{2}=4x-12$$

Subtract 4x from both sides.

$$153-45x+3x^{2}-4x=-12$$

Combine -45x and -4x to get -49x.

$$153-49x+3x^{2}=-12$$

Add 12 to both sides.

$$153-49x+3x^{2}+12=0$$

Add 153 and 12 to get 165.

$$165-49x+3x^{2}=0$$

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.

$$3x^{2}-49x+165=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -49 for b, and 165 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-49\right)±\sqrt{\left(-49\right)^{2}-4\times 3\times 165}}{2\times 3}$$

Square -49.

$$x=\frac{-\left(-49\right)±\sqrt{2401-4\times 3\times 165}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-49\right)±\sqrt{2401-12\times 165}}{2\times 3}$$

Multiply -12 times 165.

$$x=\frac{-\left(-49\right)±\sqrt{2401-1980}}{2\times 3}$$

Add 2401 to -1980.

$$x=\frac{-\left(-49\right)±\sqrt{421}}{2\times 3}$$

The opposite of -49 is 49.

$$x=\frac{49±\sqrt{421}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{49±\sqrt{421}}{6}$$

Now solve the equation x=\frac{49±\sqrt{421}}{6} when ± is plus. Add 49 to \sqrt{421}.

$$x=\frac{\sqrt{421}+49}{6}$$

Now solve the equation x=\frac{49±\sqrt{421}}{6} when ± is minus. Subtract \sqrt{421} from 49.

$$x=\frac{49-\sqrt{421}}{6}$$

The equation is now solved.

$$x=\frac{\sqrt{421}+49}{6} x=\frac{49-\sqrt{421}}{6}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(7-x\right)^{2}.

$$3\left(49-14x+x^{2}\right)-3\left(x-2\right)=4\left(x-3\right)$$

Use the distributive property to multiply 3 by 49-14x+x^{2}.

$$147-42x+3x^{2}-3\left(x-2\right)=4\left(x-3\right)$$

Use the distributive property to multiply -3 by x-2.

$$147-42x+3x^{2}-3x+6=4\left(x-3\right)$$

Combine -42x and -3x to get -45x.

$$147-45x+3x^{2}+6=4\left(x-3\right)$$

Add 147 and 6 to get 153.

$$153-45x+3x^{2}=4\left(x-3\right)$$

Use the distributive property to multiply 4 by x-3.

$$153-45x+3x^{2}=4x-12$$

Subtract 4x from both sides.

$$153-45x+3x^{2}-4x=-12$$

Combine -45x and -4x to get -49x.

$$153-49x+3x^{2}=-12$$

Subtract 153 from both sides.

$$-49x+3x^{2}=-12-153$$

Subtract 153 from -12 to get -165.

$$-49x+3x^{2}=-165$$

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.

$$3x^{2}-49x=-165$$

Divide both sides by 3.

$$\frac{3x^{2}-49x}{3}=-\frac{165}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{49}{3}x=-\frac{165}{3}$$

Divide -165 by 3.

$$x^{2}-\frac{49}{3}x=-55$$

Divide -\frac{49}{3}, the coefficient of the x term, by 2 to get -\frac{49}{6}. Then add the square of -\frac{49}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{49}{3}x+\left(-\frac{49}{6}\right)^{2}=-55+\left(-\frac{49}{6}\right)^{2}$$

Square -\frac{49}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{49}{3}x+\frac{2401}{36}=-55+\frac{2401}{36}$$

Add -55 to \frac{2401}{36}.

$$x^{2}-\frac{49}{3}x+\frac{2401}{36}=\frac{421}{36}$$

Factor x^{2}-\frac{49}{3}x+\frac{2401}{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}.

$$\left(x-\frac{49}{6}\right)^{2}=\frac{421}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{49}{6}\right)^{2}}=\sqrt{\frac{421}{36}}$$

Simplify.

$$x-\frac{49}{6}=\frac{\sqrt{421}}{6} x-\frac{49}{6}=-\frac{\sqrt{421}}{6}$$

Add \frac{49}{6} to both sides of the equation.

$$x=\frac{\sqrt{421}+49}{6} x=\frac{49-\sqrt{421}}{6}$$