Solve (x+3)(x-8)/4=(x-2)

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

Solve for x

$$x = \frac{\sqrt{145} + 9}{2} \approx 10.520797289$$

$$x=\frac{9-\sqrt{145}}{2}\approx -1.520797289$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 4.

$$\left(x+3\right)\left(x-8\right)=4x-8$$

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

$$x^{2}-5x-24=4x-8$$

Subtract 4x from both sides.

$$x^{2}-5x-24-4x=-8$$

Combine -5x and -4x to get -9x.

$$x^{2}-9x-24=-8$$

Add 8 to both sides.

$$x^{2}-9x-24+8=0$$

Add -24 and 8 to get -16.

$$x^{2}-9x-16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -9 for b, and -16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\left(-16\right)}}{2}$$

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\left(-16\right)}}{2}$$

Multiply -4 times -16.

$$x=\frac{-\left(-9\right)±\sqrt{81+64}}{2}$$

Add 81 to 64.

$$x=\frac{-\left(-9\right)±\sqrt{145}}{2}$$

The opposite of -9 is 9.

$$x=\frac{9±\sqrt{145}}{2}$$

Now solve the equation x=\frac{9±\sqrt{145}}{2} when ± is plus. Add 9 to \sqrt{145}.

$$x=\frac{\sqrt{145}+9}{2}$$

Now solve the equation x=\frac{9±\sqrt{145}}{2} when ± is minus. Subtract \sqrt{145} from 9.

$$x=\frac{9-\sqrt{145}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{145}+9}{2} x=\frac{9-\sqrt{145}}{2}$$

Steps for Completing the Square

Multiply both sides of the equation by 4.

$$\left(x+3\right)\left(x-8\right)=4x-8$$

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

$$x^{2}-5x-24=4x-8$$

Subtract 4x from both sides.

$$x^{2}-5x-24-4x=-8$$

Combine -5x and -4x to get -9x.

$$x^{2}-9x-24=-8$$

Add 24 to both sides.

$$x^{2}-9x=-8+24$$

Add -8 and 24 to get 16.

$$x^{2}-9x=16$$

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

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

Square -\frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-9x+\frac{81}{4}=16+\frac{81}{4}$$

Add 16 to \frac{81}{4}.

$$x^{2}-9x+\frac{81}{4}=\frac{145}{4}$$

Factor x^{2}-9x+\frac{81}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{9}{2}\right)^{2}=\frac{145}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{2}\right)^{2}}=\sqrt{\frac{145}{4}}$$

Simplify.

$$x-\frac{9}{2}=\frac{\sqrt{145}}{2} x-\frac{9}{2}=-\frac{\sqrt{145}}{2}$$

Add \frac{9}{2} to both sides of the equation.

$$x=\frac{\sqrt{145}+9}{2} x=\frac{9-\sqrt{145}}{2}$$