Solve left(2xright)^2-2*(-x)-3=-1

$${ \left(2x \right) }^{ 2 } -2 \times (-x)-3 = -1$$

Solve for x

$$x=-1$$

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

Steps Using Factoring By Grouping

Expand \left(2x\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

$$4x^{2}-2\left(-x\right)-3=-1$$

Add 1 to both sides.

$$4x^{2}-2\left(-x\right)-3+1=0$$

Add -3 and 1 to get -2.

$$4x^{2}-2\left(-x\right)-2=0$$

Multiply -1 and 2 to get -2.

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

Multiply -2 and -1 to get 2.

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

Divide both sides by 2.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=1 ab=2\left(-1\right)=-2$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1 b=2$$

Rewrite 2x^{2}+x-1 as \left(2x^{2}-x\right)+\left(2x-1\right).

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

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

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

Factor out common term 2x-1 by using distributive property.

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

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

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

Steps Using the Quadratic Formula

Expand \left(2x\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

$$4x^{2}-2\left(-x\right)-3=-1$$

Add 1 to both sides.

$$4x^{2}-2\left(-x\right)-3+1=0$$

Add -3 and 1 to get -2.

$$4x^{2}-2\left(-x\right)-2=0$$

Multiply -1 and 2 to get -2.

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

Multiply -2 and -1 to get 2.

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

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

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

Square 2.

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

Multiply -4 times 4.

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

Multiply -16 times -2.

$$x=\frac{-2±\sqrt{4+32}}{2\times 4}$$

Add 4 to 32.

$$x=\frac{-2±\sqrt{36}}{2\times 4}$$

Take the square root of 36.

$$x=\frac{-2±6}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-2±6}{8}$$

Now solve the equation x=\frac{-2±6}{8} when ± is plus. Add -2 to 6.

$$x=\frac{4}{8}$$

Reduce the fraction \frac{4}{8} to lowest terms by extracting and canceling out 4.

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

Now solve the equation x=\frac{-2±6}{8} when ± is minus. Subtract 6 from -2.

$$x=-\frac{8}{8}$$

Divide -8 by 8.

$$x=-1$$

The equation is now solved.

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

Steps for Completing the Square

Expand \left(2x\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

$$4x^{2}-2\left(-x\right)-3=-1$$

Add 3 to both sides.

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

Add -1 and 3 to get 2.

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

Multiply -1 and 2 to get -2.

$$4x^{2}-2\left(-1\right)x=2$$

Multiply -2 and -1 to get 2.

$$4x^{2}+2x=2$$

Divide both sides by 4.

$$\frac{4x^{2}+2x}{4}=\frac{2}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{2}{4}x=\frac{2}{4}$$

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

$$x^{2}+\frac{1}{2}x=\frac{2}{4}$$

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

$$x^{2}+\frac{1}{2}x=\frac{1}{2}$$

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=\frac{1}{2}+\left(\frac{1}{4}\right)^{2}$$

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{1}{2}+\frac{1}{16}$$

Add \frac{1}{2} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{9}{16}$$

Factor x^{2}+\frac{1}{2}x+\frac{1}{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}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

$$x+\frac{1}{4}=\frac{3}{4} x+\frac{1}{4}=-\frac{3}{4}$$

Subtract \frac{1}{4} from both sides of the equation.

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