Solve -3(2x-1)+(x+1)(x-1)-5(x+2)=1

$$- 3 ( 2 x - 1 ) + ( x + 1 ) ( x - 1 ) - 5 ( x + 2 ) = 1$$

Solve for x

$$x = \frac{\sqrt{157} + 11}{2} \approx 11.764982043$$

$$x=\frac{11-\sqrt{157}}{2}\approx -0.764982043$$

Steps Using the Quadratic Formula

Use the distributive property to multiply -3 by 2x-1.

$$-6x+3+\left(x+1\right)\left(x-1\right)-5\left(x+2\right)=1$$

Consider \left(x+1\right)\left(x-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

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

Subtract 1 from 3 to get 2.

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

Use the distributive property to multiply -5 by x+2.

$$-6x+2+x^{2}-5x-10=1$$

Combine -6x and -5x to get -11x.

$$-11x+2+x^{2}-10=1$$

Subtract 10 from 2 to get -8.

$$-11x-8+x^{2}=1$$

Subtract 1 from both sides.

$$-11x-8+x^{2}-1=0$$

Subtract 1 from -8 to get -9.

$$-11x-9+x^{2}=0$$

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.

$$x^{2}-11x-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -11 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\left(-9\right)}}{2}$$

Square -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\left(-9\right)}}{2}$$

Multiply -4 times -9.

$$x=\frac{-\left(-11\right)±\sqrt{121+36}}{2}$$

Add 121 to 36.

$$x=\frac{-\left(-11\right)±\sqrt{157}}{2}$$

The opposite of -11 is 11.

$$x=\frac{11±\sqrt{157}}{2}$$

Now solve the equation x=\frac{11±\sqrt{157}}{2} when ± is plus. Add 11 to \sqrt{157}.

$$x=\frac{\sqrt{157}+11}{2}$$

Now solve the equation x=\frac{11±\sqrt{157}}{2} when ± is minus. Subtract \sqrt{157} from 11.

$$x=\frac{11-\sqrt{157}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{157}+11}{2} x=\frac{11-\sqrt{157}}{2}$$

Steps for Completing the Square

Use the distributive property to multiply -3 by 2x-1.

$$-6x+3+\left(x+1\right)\left(x-1\right)-5\left(x+2\right)=1$$

Consider \left(x+1\right)\left(x-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

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

Subtract 1 from 3 to get 2.

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

Use the distributive property to multiply -5 by x+2.

$$-6x+2+x^{2}-5x-10=1$$

Combine -6x and -5x to get -11x.

$$-11x+2+x^{2}-10=1$$

Subtract 10 from 2 to get -8.

$$-11x-8+x^{2}=1$$

Add 8 to both sides.

$$-11x+x^{2}=1+8$$

Add 1 and 8 to get 9.

$$-11x+x^{2}=9$$

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.

$$x^{2}-11x=9$$

Divide -11, the coefficient of the x term, by 2 to get -\frac{11}{2}. Then add the square of -\frac{11}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-11x+\left(-\frac{11}{2}\right)^{2}=9+\left(-\frac{11}{2}\right)^{2}$$

Square -\frac{11}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-11x+\frac{121}{4}=9+\frac{121}{4}$$

Add 9 to \frac{121}{4}.

$$x^{2}-11x+\frac{121}{4}=\frac{157}{4}$$

Factor x^{2}-11x+\frac{121}{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}.

$$\left(x-\frac{11}{2}\right)^{2}=\frac{157}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{2}\right)^{2}}=\sqrt{\frac{157}{4}}$$

Simplify.

$$x-\frac{11}{2}=\frac{\sqrt{157}}{2} x-\frac{11}{2}=-\frac{\sqrt{157}}{2}$$

Add \frac{11}{2} to both sides of the equation.

$$x=\frac{\sqrt{157}+11}{2} x=\frac{11-\sqrt{157}}{2}$$