Solve 2(x-1)^2-7(x-1)-15=0

$$2 ( x - 1 ) ^ { 2 } - 7 ( x - 1 ) - 15 = 0$$

Solve for x

$$x=-\frac{1}{2}=-0.5$$

$$x=6$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$2\left(x^{2}-2x+1\right)-7\left(x-1\right)-15=0$$

Use the distributive property to multiply 2 by x^{2}-2x+1.

$$2x^{2}-4x+2-7\left(x-1\right)-15=0$$

Use the distributive property to multiply -7 by x-1.

$$2x^{2}-4x+2-7x+7-15=0$$

Combine -4x and -7x to get -11x.

$$2x^{2}-11x+2+7-15=0$$

Add 2 and 7 to get 9.

$$2x^{2}-11x+9-15=0$$

Subtract 15 from 9 to get -6.

$$2x^{2}-11x-6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=-11 ab=2\left(-6\right)=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -12.

$$1,-12 2,-6 3,-4$$

Calculate the sum for each pair.

$$1-12=-11 2-6=-4 3-4=-1$$

The solution is the pair that gives sum -11.

$$a=-12 b=1$$

Rewrite 2x^{2}-11x-6 as \left(2x^{2}-12x\right)+\left(x-6\right).

$$\left(2x^{2}-12x\right)+\left(x-6\right)$$

Factor out 2x in 2x^{2}-12x.

$$2x\left(x-6\right)+x-6$$

Factor out common term x-6 by using distributive property.

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

To find equation solutions, solve x-6=0 and 2x+1=0.

$$x=6 x=-\frac{1}{2}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$2\left(x^{2}-2x+1\right)-7\left(x-1\right)-15=0$$

Use the distributive property to multiply 2 by x^{2}-2x+1.

$$2x^{2}-4x+2-7\left(x-1\right)-15=0$$

Use the distributive property to multiply -7 by x-1.

$$2x^{2}-4x+2-7x+7-15=0$$

Combine -4x and -7x to get -11x.

$$2x^{2}-11x+2+7-15=0$$

Add 2 and 7 to get 9.

$$2x^{2}-11x+9-15=0$$

Subtract 15 from 9 to get -6.

$$2x^{2}-11x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -11 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 2\left(-6\right)}}{2\times 2}$$

Square -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\times 2\left(-6\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-11\right)±\sqrt{121-8\left(-6\right)}}{2\times 2}$$

Multiply -8 times -6.

$$x=\frac{-\left(-11\right)±\sqrt{121+48}}{2\times 2}$$

Add 121 to 48.

$$x=\frac{-\left(-11\right)±\sqrt{169}}{2\times 2}$$

Take the square root of 169.

$$x=\frac{-\left(-11\right)±13}{2\times 2}$$

The opposite of -11 is 11.

$$x=\frac{11±13}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{11±13}{4}$$

Now solve the equation x=\frac{11±13}{4} when ± is plus. Add 11 to 13.

$$x=\frac{24}{4}$$

Divide 24 by 4.

$$x=6$$

Now solve the equation x=\frac{11±13}{4} when ± is minus. Subtract 13 from 11.

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

Reduce the fraction \frac{-2}{4} to lowest terms by extracting and canceling out 2.

$$x=-\frac{1}{2}$$

The equation is now solved.

$$x=6 x=-\frac{1}{2}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$2\left(x^{2}-2x+1\right)-7\left(x-1\right)-15=0$$

Use the distributive property to multiply 2 by x^{2}-2x+1.

$$2x^{2}-4x+2-7\left(x-1\right)-15=0$$

Use the distributive property to multiply -7 by x-1.

$$2x^{2}-4x+2-7x+7-15=0$$

Combine -4x and -7x to get -11x.

$$2x^{2}-11x+2+7-15=0$$

Add 2 and 7 to get 9.

$$2x^{2}-11x+9-15=0$$

Subtract 15 from 9 to get -6.

$$2x^{2}-11x-6=0$$

Add 6 to both sides. Anything plus zero gives itself.

$$2x^{2}-11x=6$$

Divide both sides by 2.

$$\frac{2x^{2}-11x}{2}=\frac{6}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{11}{2}x=\frac{6}{2}$$

Divide 6 by 2.

$$x^{2}-\frac{11}{2}x=3$$

Divide -\frac{11}{2}, the coefficient of the x term, by 2 to get -\frac{11}{4}. Then add the square of -\frac{11}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{2}x+\left(-\frac{11}{4}\right)^{2}=3+\left(-\frac{11}{4}\right)^{2}$$

Square -\frac{11}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{2}x+\frac{121}{16}=3+\frac{121}{16}$$

Add 3 to \frac{121}{16}.

$$x^{2}-\frac{11}{2}x+\frac{121}{16}=\frac{169}{16}$$

Factor x^{2}-\frac{11}{2}x+\frac{121}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{11}{4}\right)^{2}=\frac{169}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{4}\right)^{2}}=\sqrt{\frac{169}{16}}$$

Simplify.

$$x-\frac{11}{4}=\frac{13}{4} x-\frac{11}{4}=-\frac{13}{4}$$

Add \frac{11}{4} to both sides of the equation.

$$x=6 x=-\frac{1}{2}$$