Solve 4x(x+1)-(x+2)(x-3)=7

$$4 x ( x + 1 ) - ( x + 2 ) ( x - 3 ) = 7$$

Solve for x

$$x=\frac{\sqrt{37}-5}{6}\approx 0.180460422$$

$$x=\frac{-\sqrt{37}-5}{6}\approx -1.847127088$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 4x by x+1.

$$4x^{2}+4x-\left(x+2\right)\left(x-3\right)=7$$

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

$$4x^{2}+4x-\left(x^{2}-x-6\right)=7$$

To find the opposite of x^{2}-x-6, find the opposite of each term.

$$4x^{2}+4x-x^{2}+x+6=7$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}+4x+x+6=7$$

Combine 4x and x to get 5x.

$$3x^{2}+5x+6=7$$

Subtract 7 from both sides.

$$3x^{2}+5x+6-7=0$$

Subtract 7 from 6 to get -1.

$$3x^{2}+5x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 5 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\times 3\left(-1\right)}}{2\times 3}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\times 3\left(-1\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-5±\sqrt{25-12\left(-1\right)}}{2\times 3}$$

Multiply -12 times -1.

$$x=\frac{-5±\sqrt{25+12}}{2\times 3}$$

Add 25 to 12.

$$x=\frac{-5±\sqrt{37}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-5±\sqrt{37}}{6}$$

Now solve the equation x=\frac{-5±\sqrt{37}}{6} when ± is plus. Add -5 to \sqrt{37}.

$$x=\frac{\sqrt{37}-5}{6}$$

Now solve the equation x=\frac{-5±\sqrt{37}}{6} when ± is minus. Subtract \sqrt{37} from -5.

$$x=\frac{-\sqrt{37}-5}{6}$$

The equation is now solved.

$$x=\frac{\sqrt{37}-5}{6} x=\frac{-\sqrt{37}-5}{6}$$

Steps for Completing the Square

Use the distributive property to multiply 4x by x+1.

$$4x^{2}+4x-\left(x+2\right)\left(x-3\right)=7$$

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

$$4x^{2}+4x-\left(x^{2}-x-6\right)=7$$

To find the opposite of x^{2}-x-6, find the opposite of each term.

$$4x^{2}+4x-x^{2}+x+6=7$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}+4x+x+6=7$$

Combine 4x and x to get 5x.

$$3x^{2}+5x+6=7$$

Subtract 6 from both sides.

$$3x^{2}+5x=7-6$$

Subtract 6 from 7 to get 1.

$$3x^{2}+5x=1$$

Divide both sides by 3.

$$\frac{3x^{2}+5x}{3}=\frac{1}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{5}{3}x=\frac{1}{3}$$

Divide \frac{5}{3}, the coefficient of the x term, by 2 to get \frac{5}{6}. Then add the square of \frac{5}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{3}x+\left(\frac{5}{6}\right)^{2}=\frac{1}{3}+\left(\frac{5}{6}\right)^{2}$$

Square \frac{5}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{3}x+\frac{25}{36}=\frac{1}{3}+\frac{25}{36}$$

Add \frac{1}{3} to \frac{25}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{5}{3}x+\frac{25}{36}=\frac{37}{36}$$

Factor x^{2}+\frac{5}{3}x+\frac{25}{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}.

$$\left(x+\frac{5}{6}\right)^{2}=\frac{37}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{6}\right)^{2}}=\sqrt{\frac{37}{36}}$$

Simplify.

$$x+\frac{5}{6}=\frac{\sqrt{37}}{6} x+\frac{5}{6}=-\frac{\sqrt{37}}{6}$$

Subtract \frac{5}{6} from both sides of the equation.

$$x=\frac{\sqrt{37}-5}{6} x=\frac{-\sqrt{37}-5}{6}$$