Solve xleft(x-2right)+x-3=4x+3

$$x \left( x-2 \right) +x-3 = 4x+3$$

Solve for x

$$x=-1$$

$$x=6$$

Steps Using the Quadratic Formula

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

$$x^{2}-2x+x-3=4x+3$$

Combine -2x and x to get -x.

$$x^{2}-x-3=4x+3$$

Subtract 4x from both sides.

$$x^{2}-x-3-4x=3$$

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

$$x^{2}-5x-3=3$$

Subtract 3 from both sides.

$$x^{2}-5x-3-3=0$$

Subtract 3 from -3 to get -6.

$$x^{2}-5x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -5 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-6\right)}}{2}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-6\right)}}{2}$$

Multiply -4 times -6.

$$x=\frac{-\left(-5\right)±\sqrt{25+24}}{2}$$

Add 25 to 24.

$$x=\frac{-\left(-5\right)±\sqrt{49}}{2}$$

Take the square root of 49.

$$x=\frac{-\left(-5\right)±7}{2}$$

The opposite of -5 is 5.

$$x=\frac{5±7}{2}$$

Now solve the equation x=\frac{5±7}{2} when ± is plus. Add 5 to 7.

$$x=\frac{12}{2}$$

Divide 12 by 2.

$$x=6$$

Now solve the equation x=\frac{5±7}{2} when ± is minus. Subtract 7 from 5.

$$x=-\frac{2}{2}$$

Divide -2 by 2.

$$x=-1$$

The equation is now solved.

$$x=6 x=-1$$

Steps for Completing the Square

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

$$x^{2}-2x+x-3=4x+3$$

Combine -2x and x to get -x.

$$x^{2}-x-3=4x+3$$

Subtract 4x from both sides.

$$x^{2}-x-3-4x=3$$

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

$$x^{2}-5x-3=3$$

Add 3 to both sides.

$$x^{2}-5x=3+3$$

Add 3 and 3 to get 6.

$$x^{2}-5x=6$$

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=6+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=6+\frac{25}{4}$$

Add 6 to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{49}{4}$$

Factor x^{2}-5x+\frac{25}{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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{7}{2} x-\frac{5}{2}=-\frac{7}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$x=6 x=-1$$