Solve -1=4x^2+2x-4(1-x^2)

$$-1=4 { x }^{ 2 } +2x-4(1- { x }^{ 2 } )$$

Solve for x

$$x=-\frac{3}{4}=-0.75$$

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

Steps Using Factoring By Grouping

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

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

Combine 4x^{2} and 4x^{2} to get 8x^{2}.

$$-1=8x^{2}+2x-4$$

Swap sides so that all variable terms are on the left hand side.

$$8x^{2}+2x-4=-1$$

Add 1 to both sides.

$$8x^{2}+2x-4+1=0$$

Add -4 and 1 to get -3.

$$8x^{2}+2x-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 8x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=2 ab=8\left(-3\right)=-24$$

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. List all such integer pairs that give product -24.

$$-1,24 -2,12 -3,8 -4,6$$

Calculate the sum for each pair.

$$-1+24=23 -2+12=10 -3+8=5 -4+6=2$$

The solution is the pair that gives sum 2.

$$a=-4 b=6$$

Rewrite 8x^{2}+2x-3 as \left(8x^{2}-4x\right)+\left(6x-3\right).

$$\left(8x^{2}-4x\right)+\left(6x-3\right)$$

Factor out 4x in the first and 3 in the second group.

$$4x\left(2x-1\right)+3\left(2x-1\right)$$

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

$$\left(2x-1\right)\left(4x+3\right)$$

To find equation solutions, solve 2x-1=0 and 4x+3=0.

$$x=\frac{1}{2} x=-\frac{3}{4}$$

Steps Using the Quadratic Formula

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

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

Combine 4x^{2} and 4x^{2} to get 8x^{2}.

$$-1=8x^{2}+2x-4$$

Swap sides so that all variable terms are on the left hand side.

$$8x^{2}+2x-4=-1$$

Add 1 to both sides.

$$8x^{2}+2x-4+1=0$$

Add -4 and 1 to get -3.

$$8x^{2}+2x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, 2 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\times 8\left(-3\right)}}{2\times 8}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\times 8\left(-3\right)}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-2±\sqrt{4-32\left(-3\right)}}{2\times 8}$$

Multiply -32 times -3.

$$x=\frac{-2±\sqrt{4+96}}{2\times 8}$$

Add 4 to 96.

$$x=\frac{-2±\sqrt{100}}{2\times 8}$$

Take the square root of 100.

$$x=\frac{-2±10}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{-2±10}{16}$$

Now solve the equation x=\frac{-2±10}{16} when ± is plus. Add -2 to 10.

$$x=\frac{8}{16}$$

Reduce the fraction \frac{8}{16} to lowest terms by extracting and canceling out 8.

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

Now solve the equation x=\frac{-2±10}{16} when ± is minus. Subtract 10 from -2.

$$x=-\frac{12}{16}$$

Reduce the fraction \frac{-12}{16} to lowest terms by extracting and canceling out 4.

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

The equation is now solved.

$$x=\frac{1}{2} x=-\frac{3}{4}$$

Steps for Completing the Square

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

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

Combine 4x^{2} and 4x^{2} to get 8x^{2}.

$$-1=8x^{2}+2x-4$$

Swap sides so that all variable terms are on the left hand side.

$$8x^{2}+2x-4=-1$$

Add 4 to both sides.

$$8x^{2}+2x=-1+4$$

Add -1 and 4 to get 3.

$$8x^{2}+2x=3$$

Divide both sides by 8.

$$\frac{8x^{2}+2x}{8}=\frac{3}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x^{2}+\frac{2}{8}x=\frac{3}{8}$$

Reduce the fraction \frac{2}{8} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{1}{4}x=\frac{3}{8}$$

Divide \frac{1}{4}, the coefficient of the x term, by 2 to get \frac{1}{8}. Then add the square of \frac{1}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{4}x+\left(\frac{1}{8}\right)^{2}=\frac{3}{8}+\left(\frac{1}{8}\right)^{2}$$

Square \frac{1}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{4}x+\frac{1}{64}=\frac{3}{8}+\frac{1}{64}$$

Add \frac{3}{8} to \frac{1}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{4}x+\frac{1}{64}=\frac{25}{64}$$

Factor x^{2}+\frac{1}{4}x+\frac{1}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{8}\right)^{2}=\frac{25}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{8}\right)^{2}}=\sqrt{\frac{25}{64}}$$

Simplify.

$$x+\frac{1}{8}=\frac{5}{8} x+\frac{1}{8}=-\frac{5}{8}$$

Subtract \frac{1}{8} from both sides of the equation.

$$x=\frac{1}{2} x=-\frac{3}{4}$$