Solve 4=(4x^2-1)+(x-3)

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

Solve for x

$$x = \frac{\sqrt{129} - 1}{8} \approx 1.294727086$$

$$x=\frac{-\sqrt{129}-1}{8}\approx -1.544727086$$

Steps Using the Quadratic Formula

Subtract 3 from -1 to get -4.

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

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

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

Subtract 4 from both sides.

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

Subtract 4 from -4 to get -8.

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 1 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\times 4\left(-8\right)}}{2\times 4}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\times 4\left(-8\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-1±\sqrt{1-16\left(-8\right)}}{2\times 4}$$

Multiply -16 times -8.

$$x=\frac{-1±\sqrt{1+128}}{2\times 4}$$

Add 1 to 128.

$$x=\frac{-1±\sqrt{129}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-1±\sqrt{129}}{8}$$

Now solve the equation x=\frac{-1±\sqrt{129}}{8} when ± is plus. Add -1 to \sqrt{129}.

$$x=\frac{\sqrt{129}-1}{8}$$

Now solve the equation x=\frac{-1±\sqrt{129}}{8} when ± is minus. Subtract \sqrt{129} from -1.

$$x=\frac{-\sqrt{129}-1}{8}$$

The equation is now solved.

$$x=\frac{\sqrt{129}-1}{8} x=\frac{-\sqrt{129}-1}{8}$$

Steps for Completing the Square

Subtract 3 from -1 to get -4.

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

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

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

Add 4 to both sides.

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

Add 4 and 4 to get 8.

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

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

Divide 8 by 4.

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

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.

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

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

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

Add 2 to \frac{1}{64}.

$$x^{2}+\frac{1}{4}x+\frac{1}{64}=\frac{129}{64}$$

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}.

$$\left(x+\frac{1}{8}\right)^{2}=\frac{129}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{8}\right)^{2}}=\sqrt{\frac{129}{64}}$$

Simplify.

$$x+\frac{1}{8}=\frac{\sqrt{129}}{8} x+\frac{1}{8}=-\frac{\sqrt{129}}{8}$$

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

$$x=\frac{\sqrt{129}-1}{8} x=\frac{-\sqrt{129}-1}{8}$$