Solve 10x-8x-3/4/x-2x-3=2(x-3)

$$\frac { 10 x - \frac { 8 x - 3 } { 4 } } { x - 2 x - 3 } = 2 ( x - 3 )$$

Solve for x

$$x=\frac{\sqrt{202}}{4}-2\approx 1.553167601$$

$$x=-\frac{\sqrt{202}}{4}-2\approx -5.553167601$$

Steps Using the Quadratic Formula

Variable x cannot be equal to -3 since division by zero is not defined. Multiply both sides of the equation by -x-3.

$$10x-\frac{8x-3}{4}=2\left(x-3\right)\left(-x-3\right)$$

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

$$10x-\frac{8x-3}{4}=\left(2x-6\right)\left(-x-3\right)$$

Use the distributive property to multiply 2x-6 by -x-3 and combine like terms.

$$10x-\frac{8x-3}{4}=-2x^{2}+18$$

Divide each term of 8x-3 by 4 to get 2x-\frac{3}{4}.

$$10x-\left(2x-\frac{3}{4}\right)=-2x^{2}+18$$

To find the opposite of 2x-\frac{3}{4}, find the opposite of each term.

$$10x-2x+\frac{3}{4}=-2x^{2}+18$$

Combine 10x and -2x to get 8x.

$$8x+\frac{3}{4}=-2x^{2}+18$$

Add 2x^{2} to both sides.

$$8x+\frac{3}{4}+2x^{2}=18$$

Subtract 18 from both sides.

$$8x+\frac{3}{4}+2x^{2}-18=0$$

Subtract 18 from \frac{3}{4} to get -\frac{69}{4}.

$$8x-\frac{69}{4}+2x^{2}=0$$

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.

$$2x^{2}+8x-\frac{69}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 8 for b, and -\frac{69}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\times 2\left(-\frac{69}{4}\right)}}{2\times 2}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 2\left(-\frac{69}{4}\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-8±\sqrt{64-8\left(-\frac{69}{4}\right)}}{2\times 2}$$

Multiply -8 times -\frac{69}{4}.

$$x=\frac{-8±\sqrt{64+138}}{2\times 2}$$

Add 64 to 138.

$$x=\frac{-8±\sqrt{202}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-8±\sqrt{202}}{4}$$

Now solve the equation x=\frac{-8±\sqrt{202}}{4} when ± is plus. Add -8 to \sqrt{202}.

$$x=\frac{\sqrt{202}-8}{4}$$

Divide -8+\sqrt{202} by 4.

$$x=\frac{\sqrt{202}}{4}-2$$

Now solve the equation x=\frac{-8±\sqrt{202}}{4} when ± is minus. Subtract \sqrt{202} from -8.

$$x=\frac{-\sqrt{202}-8}{4}$$

Divide -8-\sqrt{202} by 4.

$$x=-\frac{\sqrt{202}}{4}-2$$

The equation is now solved.

$$x=\frac{\sqrt{202}}{4}-2 x=-\frac{\sqrt{202}}{4}-2$$

Steps for Completing the Square

Variable x cannot be equal to -3 since division by zero is not defined. Multiply both sides of the equation by -x-3.

$$10x-\frac{8x-3}{4}=2\left(x-3\right)\left(-x-3\right)$$

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

$$10x-\frac{8x-3}{4}=\left(2x-6\right)\left(-x-3\right)$$

Use the distributive property to multiply 2x-6 by -x-3 and combine like terms.

$$10x-\frac{8x-3}{4}=-2x^{2}+18$$

Divide each term of 8x-3 by 4 to get 2x-\frac{3}{4}.

$$10x-\left(2x-\frac{3}{4}\right)=-2x^{2}+18$$

To find the opposite of 2x-\frac{3}{4}, find the opposite of each term.

$$10x-2x+\frac{3}{4}=-2x^{2}+18$$

Combine 10x and -2x to get 8x.

$$8x+\frac{3}{4}=-2x^{2}+18$$

Add 2x^{2} to both sides.

$$8x+\frac{3}{4}+2x^{2}=18$$

Subtract \frac{3}{4} from both sides.

$$8x+2x^{2}=18-\frac{3}{4}$$

Subtract \frac{3}{4} from 18 to get \frac{69}{4}.

$$8x+2x^{2}=\frac{69}{4}$$

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.

$$2x^{2}+8x=\frac{69}{4}$$

Divide both sides by 2.

$$\frac{2x^{2}+8x}{2}=\frac{\frac{69}{4}}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{8}{2}x=\frac{\frac{69}{4}}{2}$$

Divide 8 by 2.

$$x^{2}+4x=\frac{\frac{69}{4}}{2}$$

Divide \frac{69}{4} by 2.

$$x^{2}+4x=\frac{69}{8}$$

Divide 4, the coefficient of the x term, by 2 to get 2. Then add the square of 2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+4x+2^{2}=\frac{69}{8}+2^{2}$$

Square 2.

$$x^{2}+4x+4=\frac{69}{8}+4$$

Add \frac{69}{8} to 4.

$$x^{2}+4x+4=\frac{101}{8}$$

Factor x^{2}+4x+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}.

$$\left(x+2\right)^{2}=\frac{101}{8}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{\frac{101}{8}}$$

Simplify.

$$x+2=\frac{\sqrt{202}}{4} x+2=-\frac{\sqrt{202}}{4}$$

Subtract 2 from both sides of the equation.

$$x=\frac{\sqrt{202}}{4}-2 x=-\frac{\sqrt{202}}{4}-2$$