Solve 2(x+1)(x-1)/3-5x-3/2=-(2x+1)^2/6+7/4

$$\frac { 2 ( x + 1 ) ( x - 1 ) } { 3 } - \frac { 5 x - 3 } { 2 } = - \frac { ( 2 x + 1 ) ^ { 2 } } { 6 } + \frac { 7 } { 4 }$$

Solve for x

$$x = \frac{\sqrt{265} + 11}{16} \approx 1.704926287$$

$$x=\frac{11-\sqrt{265}}{16}\approx -0.329926287$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 12, the least common multiple of 3,2,6,4.

$$4\times 2\left(x+1\right)\left(x-1\right)-6\left(5x-3\right)=-2\left(2x+1\right)^{2}+21$$

Multiply 4 and 2 to get 8.

$$8\left(x+1\right)\left(x-1\right)-6\left(5x-3\right)=-2\left(2x+1\right)^{2}+21$$

Use the distributive property to multiply 8 by x+1.

$$\left(8x+8\right)\left(x-1\right)-6\left(5x-3\right)=-2\left(2x+1\right)^{2}+21$$

Use the distributive property to multiply 8x+8 by x-1 and combine like terms.

$$8x^{2}-8-6\left(5x-3\right)=-2\left(2x+1\right)^{2}+21$$

Use the distributive property to multiply -6 by 5x-3.

$$8x^{2}-8-30x+18=-2\left(2x+1\right)^{2}+21$$

Add -8 and 18 to get 10.

$$8x^{2}+10-30x=-2\left(2x+1\right)^{2}+21$$

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

$$8x^{2}+10-30x=-2\left(4x^{2}+4x+1\right)+21$$

Use the distributive property to multiply -2 by 4x^{2}+4x+1.

$$8x^{2}+10-30x=-8x^{2}-8x-2+21$$

Add -2 and 21 to get 19.

$$8x^{2}+10-30x=-8x^{2}-8x+19$$

Add 8x^{2} to both sides.

$$8x^{2}+10-30x+8x^{2}=-8x+19$$

Combine 8x^{2} and 8x^{2} to get 16x^{2}.

$$16x^{2}+10-30x=-8x+19$$

Add 8x to both sides.

$$16x^{2}+10-30x+8x=19$$

Combine -30x and 8x to get -22x.

$$16x^{2}+10-22x=19$$

Subtract 19 from both sides.

$$16x^{2}+10-22x-19=0$$

Subtract 19 from 10 to get -9.

$$16x^{2}-9-22x=0$$

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.

$$16x^{2}-22x-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, -22 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-22\right)±\sqrt{\left(-22\right)^{2}-4\times 16\left(-9\right)}}{2\times 16}$$

Square -22.

$$x=\frac{-\left(-22\right)±\sqrt{484-4\times 16\left(-9\right)}}{2\times 16}$$

Multiply -4 times 16.

$$x=\frac{-\left(-22\right)±\sqrt{484-64\left(-9\right)}}{2\times 16}$$

Multiply -64 times -9.

$$x=\frac{-\left(-22\right)±\sqrt{484+576}}{2\times 16}$$

Add 484 to 576.

$$x=\frac{-\left(-22\right)±\sqrt{1060}}{2\times 16}$$

Take the square root of 1060.

$$x=\frac{-\left(-22\right)±2\sqrt{265}}{2\times 16}$$

The opposite of -22 is 22.

$$x=\frac{22±2\sqrt{265}}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{22±2\sqrt{265}}{32}$$

Now solve the equation x=\frac{22±2\sqrt{265}}{32} when ± is plus. Add 22 to 2\sqrt{265}.

$$x=\frac{2\sqrt{265}+22}{32}$$

Divide 22+2\sqrt{265} by 32.

$$x=\frac{\sqrt{265}+11}{16}$$

Now solve the equation x=\frac{22±2\sqrt{265}}{32} when ± is minus. Subtract 2\sqrt{265} from 22.

$$x=\frac{22-2\sqrt{265}}{32}$$

Divide 22-2\sqrt{265} by 32.

$$x=\frac{11-\sqrt{265}}{16}$$

The equation is now solved.

$$x=\frac{\sqrt{265}+11}{16} x=\frac{11-\sqrt{265}}{16}$$

Steps for Completing the Square

Multiply both sides of the equation by 12, the least common multiple of 3,2,6,4.

$$4\times 2\left(x+1\right)\left(x-1\right)-6\left(5x-3\right)=-2\left(2x+1\right)^{2}+21$$

Multiply 4 and 2 to get 8.

$$8\left(x+1\right)\left(x-1\right)-6\left(5x-3\right)=-2\left(2x+1\right)^{2}+21$$

Use the distributive property to multiply 8 by x+1.

$$\left(8x+8\right)\left(x-1\right)-6\left(5x-3\right)=-2\left(2x+1\right)^{2}+21$$

Use the distributive property to multiply 8x+8 by x-1 and combine like terms.

$$8x^{2}-8-6\left(5x-3\right)=-2\left(2x+1\right)^{2}+21$$

Use the distributive property to multiply -6 by 5x-3.

$$8x^{2}-8-30x+18=-2\left(2x+1\right)^{2}+21$$

Add -8 and 18 to get 10.

$$8x^{2}+10-30x=-2\left(2x+1\right)^{2}+21$$

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

$$8x^{2}+10-30x=-2\left(4x^{2}+4x+1\right)+21$$

Use the distributive property to multiply -2 by 4x^{2}+4x+1.

$$8x^{2}+10-30x=-8x^{2}-8x-2+21$$

Add -2 and 21 to get 19.

$$8x^{2}+10-30x=-8x^{2}-8x+19$$

Add 8x^{2} to both sides.

$$8x^{2}+10-30x+8x^{2}=-8x+19$$

Combine 8x^{2} and 8x^{2} to get 16x^{2}.

$$16x^{2}+10-30x=-8x+19$$

Add 8x to both sides.

$$16x^{2}+10-30x+8x=19$$

Combine -30x and 8x to get -22x.

$$16x^{2}+10-22x=19$$

Subtract 10 from both sides.

$$16x^{2}-22x=19-10$$

Subtract 10 from 19 to get 9.

$$16x^{2}-22x=9$$

Divide both sides by 16.

$$\frac{16x^{2}-22x}{16}=\frac{9}{16}$$

Dividing by 16 undoes the multiplication by 16.

$$x^{2}+\left(-\frac{22}{16}\right)x=\frac{9}{16}$$

Reduce the fraction \frac{-22}{16} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{11}{8}x=\frac{9}{16}$$

Divide -\frac{11}{8}, the coefficient of the x term, by 2 to get -\frac{11}{16}. Then add the square of -\frac{11}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{8}x+\left(-\frac{11}{16}\right)^{2}=\frac{9}{16}+\left(-\frac{11}{16}\right)^{2}$$

Square -\frac{11}{16} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{8}x+\frac{121}{256}=\frac{9}{16}+\frac{121}{256}$$

Add \frac{9}{16} to \frac{121}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{11}{8}x+\frac{121}{256}=\frac{265}{256}$$

Factor x^{2}-\frac{11}{8}x+\frac{121}{256}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{11}{16}\right)^{2}=\frac{265}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{16}\right)^{2}}=\sqrt{\frac{265}{256}}$$

Simplify.

$$x-\frac{11}{16}=\frac{\sqrt{265}}{16} x-\frac{11}{16}=-\frac{\sqrt{265}}{16}$$

Add \frac{11}{16} to both sides of the equation.

$$x=\frac{\sqrt{265}+11}{16} x=\frac{11-\sqrt{265}}{16}$$