Solve (x)=2(x-3)^2-(3-x)(x+8)

$$( x ) = 2 ( x - 3 ) ^ { 2 } - ( 3 - x ) ( x + 8 )$$

Solve for x

$$x = \frac{\sqrt{34} + 4}{3} \approx 3.276983965$$

$$x=\frac{4-\sqrt{34}}{3}\approx -0.610317298$$

Steps Using the Quadratic Formula

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

$$x=2\left(x^{2}-6x+9\right)-\left(3-x\right)\left(x+8\right)$$

Use the distributive property to multiply 2 by x^{2}-6x+9.

$$x=2x^{2}-12x+18-\left(3-x\right)\left(x+8\right)$$

Use the distributive property to multiply 3-x by x+8 and combine like terms.

$$x=2x^{2}-12x+18-\left(-5x+24-x^{2}\right)$$

To find the opposite of -5x+24-x^{2}, find the opposite of each term.

$$x=2x^{2}-12x+18+5x-24+x^{2}$$

Combine -12x and 5x to get -7x.

$$x=2x^{2}-7x+18-24+x^{2}$$

Subtract 24 from 18 to get -6.

$$x=2x^{2}-7x-6+x^{2}$$

Combine 2x^{2} and x^{2} to get 3x^{2}.

$$x=3x^{2}-7x-6$$

Subtract 3x^{2} from both sides.

$$x-3x^{2}=-7x-6$$

Add 7x to both sides.

$$x-3x^{2}+7x=-6$$

Combine x and 7x to get 8x.

$$8x-3x^{2}=-6$$

Add 6 to both sides.

$$8x-3x^{2}+6=0$$

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.

$$-3x^{2}+8x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 8 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\left(-3\right)\times 6}}{2\left(-3\right)}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\left(-3\right)\times 6}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-8±\sqrt{64+12\times 6}}{2\left(-3\right)}$$

Multiply 12 times 6.

$$x=\frac{-8±\sqrt{64+72}}{2\left(-3\right)}$$

Add 64 to 72.

$$x=\frac{-8±\sqrt{136}}{2\left(-3\right)}$$

Take the square root of 136.

$$x=\frac{-8±2\sqrt{34}}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-8±2\sqrt{34}}{-6}$$

Now solve the equation x=\frac{-8±2\sqrt{34}}{-6} when ± is plus. Add -8 to 2\sqrt{34}.

$$x=\frac{2\sqrt{34}-8}{-6}$$

Divide -8+2\sqrt{34} by -6.

$$x=\frac{4-\sqrt{34}}{3}$$

Now solve the equation x=\frac{-8±2\sqrt{34}}{-6} when ± is minus. Subtract 2\sqrt{34} from -8.

$$x=\frac{-2\sqrt{34}-8}{-6}$$

Divide -8-2\sqrt{34} by -6.

$$x=\frac{\sqrt{34}+4}{3}$$

The equation is now solved.

$$x=\frac{4-\sqrt{34}}{3} x=\frac{\sqrt{34}+4}{3}$$

Steps for Completing the Square

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

$$x=2\left(x^{2}-6x+9\right)-\left(3-x\right)\left(x+8\right)$$

Use the distributive property to multiply 2 by x^{2}-6x+9.

$$x=2x^{2}-12x+18-\left(3-x\right)\left(x+8\right)$$

Use the distributive property to multiply 3-x by x+8 and combine like terms.

$$x=2x^{2}-12x+18-\left(-5x+24-x^{2}\right)$$

To find the opposite of -5x+24-x^{2}, find the opposite of each term.

$$x=2x^{2}-12x+18+5x-24+x^{2}$$

Combine -12x and 5x to get -7x.

$$x=2x^{2}-7x+18-24+x^{2}$$

Subtract 24 from 18 to get -6.

$$x=2x^{2}-7x-6+x^{2}$$

Combine 2x^{2} and x^{2} to get 3x^{2}.

$$x=3x^{2}-7x-6$$

Subtract 3x^{2} from both sides.

$$x-3x^{2}=-7x-6$$

Add 7x to both sides.

$$x-3x^{2}+7x=-6$$

Combine x and 7x to get 8x.

$$8x-3x^{2}=-6$$

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.

$$-3x^{2}+8x=-6$$

Divide both sides by -3.

$$\frac{-3x^{2}+8x}{-3}=-\frac{6}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{8}{-3}x=-\frac{6}{-3}$$

Divide 8 by -3.

$$x^{2}-\frac{8}{3}x=-\frac{6}{-3}$$

Divide -6 by -3.

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

Divide -\frac{8}{3}, the coefficient of the x term, by 2 to get -\frac{4}{3}. Then add the square of -\frac{4}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{4}{3} by squaring both the numerator and the denominator of the fraction.

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

Add 2 to \frac{16}{9}.

$$x^{2}-\frac{8}{3}x+\frac{16}{9}=\frac{34}{9}$$

Factor x^{2}-\frac{8}{3}x+\frac{16}{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}.

$$\left(x-\frac{4}{3}\right)^{2}=\frac{34}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{4}{3}\right)^{2}}=\sqrt{\frac{34}{9}}$$

Simplify.

$$x-\frac{4}{3}=\frac{\sqrt{34}}{3} x-\frac{4}{3}=-\frac{\sqrt{34}}{3}$$

Add \frac{4}{3} to both sides of the equation.

$$x=\frac{\sqrt{34}+4}{3} x=\frac{4-\sqrt{34}}{3}$$