Solve (x+3)(x+2)+5x-6=6

$$(x+3)(x+2)+5x-6=6$$

Solve for x (complex solution)

$$x=\sqrt{31}-5\approx 0.567764363$$

$$x=-\left(\sqrt{31}+5\right)\approx -10.567764363$$

Steps Using the Quadratic Formula

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

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

Combine 5x and 5x to get 10x.

$$x^{2}+10x+6-6=6$$

Subtract 6 from 6 to get 0.

$$x^{2}+10x=6$$

Subtract 6 from both sides.

$$x^{2}+10x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 10 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-6\right)}}{2}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\left(-6\right)}}{2}$$

Multiply -4 times -6.

$$x=\frac{-10±\sqrt{100+24}}{2}$$

Add 100 to 24.

$$x=\frac{-10±\sqrt{124}}{2}$$

Take the square root of 124.

$$x=\frac{-10±2\sqrt{31}}{2}$$

Now solve the equation x=\frac{-10±2\sqrt{31}}{2} when ± is plus. Add -10 to 2\sqrt{31}.

$$x=\frac{2\sqrt{31}-10}{2}$$

Divide -10+2\sqrt{31} by 2.

$$x=\sqrt{31}-5$$

Now solve the equation x=\frac{-10±2\sqrt{31}}{2} when ± is minus. Subtract 2\sqrt{31} from -10.

$$x=\frac{-2\sqrt{31}-10}{2}$$

Divide -10-2\sqrt{31} by 2.

$$x=-\sqrt{31}-5$$

The equation is now solved.

$$x=\sqrt{31}-5 x=-\sqrt{31}-5$$

Steps for Completing the Square

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

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

Combine 5x and 5x to get 10x.

$$x^{2}+10x+6-6=6$$

Subtract 6 from 6 to get 0.

$$x^{2}+10x=6$$

Divide 10, the coefficient of the x term, by 2 to get 5. Then add the square of 5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+10x+5^{2}=6+5^{2}$$

Square 5.

$$x^{2}+10x+25=6+25$$

Add 6 to 25.

$$x^{2}+10x+25=31$$

Factor x^{2}+10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+5\right)^{2}=31$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{31}$$

Simplify.

$$x+5=\sqrt{31} x+5=-\sqrt{31}$$

Subtract 5 from both sides of the equation.

$$x=\sqrt{31}-5 x=-\sqrt{31}-5$$

Solve for x

$$x=\sqrt{31}-5\approx 0.567764363$$

$$x=-\sqrt{31}-5\approx -10.567764363$$

Steps Using the Quadratic Formula

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

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

Combine 5x and 5x to get 10x.

$$x^{2}+10x+6-6=6$$

Subtract 6 from 6 to get 0.

$$x^{2}+10x=6$$

Subtract 6 from both sides.

$$x^{2}+10x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 10 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-6\right)}}{2}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\left(-6\right)}}{2}$$

Multiply -4 times -6.

$$x=\frac{-10±\sqrt{100+24}}{2}$$

Add 100 to 24.

$$x=\frac{-10±\sqrt{124}}{2}$$

Take the square root of 124.

$$x=\frac{-10±2\sqrt{31}}{2}$$

Now solve the equation x=\frac{-10±2\sqrt{31}}{2} when ± is plus. Add -10 to 2\sqrt{31}.

$$x=\frac{2\sqrt{31}-10}{2}$$

Divide -10+2\sqrt{31} by 2.

$$x=\sqrt{31}-5$$

Now solve the equation x=\frac{-10±2\sqrt{31}}{2} when ± is minus. Subtract 2\sqrt{31} from -10.

$$x=\frac{-2\sqrt{31}-10}{2}$$

Divide -10-2\sqrt{31} by 2.

$$x=-\sqrt{31}-5$$

The equation is now solved.

$$x=\sqrt{31}-5 x=-\sqrt{31}-5$$

Steps for Completing the Square

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

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

Combine 5x and 5x to get 10x.

$$x^{2}+10x+6-6=6$$

Subtract 6 from 6 to get 0.

$$x^{2}+10x=6$$

Divide 10, the coefficient of the x term, by 2 to get 5. Then add the square of 5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+10x+5^{2}=6+5^{2}$$

Square 5.

$$x^{2}+10x+25=6+25$$

Add 6 to 25.

$$x^{2}+10x+25=31$$

Factor x^{2}+10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+5\right)^{2}=31$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{31}$$

Simplify.

$$x+5=\sqrt{31} x+5=-\sqrt{31}$$

Subtract 5 from both sides of the equation.

$$x=\sqrt{31}-5 x=-\sqrt{31}-5$$