Solve 2x^2-1=left(x+2right)^2+6x-14

$$2 { x }^{ 2 } -1= { \left(x+2 \right) }^{ 2 } +6x-14$$

Solve for x

$$x=1$$

$$x=9$$

Steps Using Factoring

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

$$2x^{2}-1=x^{2}+4x+4+6x-14$$

Combine 4x and 6x to get 10x.

$$2x^{2}-1=x^{2}+10x+4-14$$

Subtract 14 from 4 to get -10.

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

Subtract x^{2} from both sides.

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

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

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

Subtract 10x from both sides.

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

Add 10 to both sides.

$$x^{2}-1-10x+10=0$$

Add -1 and 10 to get 9.

$$x^{2}+9-10x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}-10x+9=0$$

To solve the equation, factor x^{2}-10x+9 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-10 ab=9$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 9.

$$-1,-9 -3,-3$$

Calculate the sum for each pair.

$$-1-9=-10 -3-3=-6$$

The solution is the pair that gives sum -10.

$$a=-9 b=-1$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-9\right)\left(x-1\right)$$

To find equation solutions, solve x-9=0 and x-1=0.

$$x=9 x=1$$

Steps Using Factoring By Grouping

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

$$2x^{2}-1=x^{2}+4x+4+6x-14$$

Combine 4x and 6x to get 10x.

$$2x^{2}-1=x^{2}+10x+4-14$$

Subtract 14 from 4 to get -10.

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

Subtract x^{2} from both sides.

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

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

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

Subtract 10x from both sides.

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

Add 10 to both sides.

$$x^{2}-1-10x+10=0$$

Add -1 and 10 to get 9.

$$x^{2}+9-10x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}-10x+9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+9. To find a and b, set up a system to be solved.

$$a+b=-10 ab=1\times 9=9$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 9.

$$-1,-9 -3,-3$$

Calculate the sum for each pair.

$$-1-9=-10 -3-3=-6$$

The solution is the pair that gives sum -10.

$$a=-9 b=-1$$

Rewrite x^{2}-10x+9 as \left(x^{2}-9x\right)+\left(-x+9\right).

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

Factor out x in the first and -1 in the second group.

$$x\left(x-9\right)-\left(x-9\right)$$

Factor out common term x-9 by using distributive property.

$$\left(x-9\right)\left(x-1\right)$$

To find equation solutions, solve x-9=0 and x-1=0.

$$x=9 x=1$$

Steps Using the Quadratic Formula

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

$$2x^{2}-1=x^{2}+4x+4+6x-14$$

Combine 4x and 6x to get 10x.

$$2x^{2}-1=x^{2}+10x+4-14$$

Subtract 14 from 4 to get -10.

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

Subtract x^{2} from both sides.

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

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

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

Subtract 10x from both sides.

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

Add 10 to both sides.

$$x^{2}-1-10x+10=0$$

Add -1 and 10 to get 9.

$$x^{2}+9-10x=0$$

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.

$$x^{2}-10x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 9}}{2}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 9}}{2}$$

Multiply -4 times 9.

$$x=\frac{-\left(-10\right)±\sqrt{100-36}}{2}$$

Add 100 to -36.

$$x=\frac{-\left(-10\right)±\sqrt{64}}{2}$$

Take the square root of 64.

$$x=\frac{-\left(-10\right)±8}{2}$$

The opposite of -10 is 10.

$$x=\frac{10±8}{2}$$

Now solve the equation x=\frac{10±8}{2} when ± is plus. Add 10 to 8.

$$x=\frac{18}{2}$$

Divide 18 by 2.

$$x=9$$

Now solve the equation x=\frac{10±8}{2} when ± is minus. Subtract 8 from 10.

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

Divide 2 by 2.

$$x=1$$

The equation is now solved.

$$x=9 x=1$$

Steps for Completing the Square

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

$$2x^{2}-1=x^{2}+4x+4+6x-14$$

Combine 4x and 6x to get 10x.

$$2x^{2}-1=x^{2}+10x+4-14$$

Subtract 14 from 4 to get -10.

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

Subtract x^{2} from both sides.

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

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

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

Subtract 10x from both sides.

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

Add 1 to both sides.

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

Add -10 and 1 to get -9.

$$x^{2}-10x=-9$$

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.

$$x^{2}-10x+\left(-5\right)^{2}=-9+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=-9+25$$

Add -9 to 25.

$$x^{2}-10x+25=16$$

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}.

$$\left(x-5\right)^{2}=16$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{16}$$

Simplify.

$$x-5=4 x-5=-4$$

Add 5 to both sides of the equation.

$$x=9 x=1$$