Solve 4x^2+82x=4x+8

$$4 { x }^{ 2 } +82x=4x+8$$

Solve for x

$$x=\frac{\sqrt{1553}-39}{4}\approx 0.102030248$$

$$x=\frac{-\sqrt{1553}-39}{4}\approx -19.602030248$$

Steps Using the Quadratic Formula

Subtract 4x from both sides.

$$4x^{2}+82x-4x=8$$

Combine 82x and -4x to get 78x.

$$4x^{2}+78x=8$$

Subtract 8 from both sides.

$$4x^{2}+78x-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 78 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-78±\sqrt{78^{2}-4\times 4\left(-8\right)}}{2\times 4}$$

Square 78.

$$x=\frac{-78±\sqrt{6084-4\times 4\left(-8\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-78±\sqrt{6084-16\left(-8\right)}}{2\times 4}$$

Multiply -16 times -8.

$$x=\frac{-78±\sqrt{6084+128}}{2\times 4}$$

Add 6084 to 128.

$$x=\frac{-78±\sqrt{6212}}{2\times 4}$$

Take the square root of 6212.

$$x=\frac{-78±2\sqrt{1553}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-78±2\sqrt{1553}}{8}$$

Now solve the equation x=\frac{-78±2\sqrt{1553}}{8} when ± is plus. Add -78 to 2\sqrt{1553}.

$$x=\frac{2\sqrt{1553}-78}{8}$$

Divide -78+2\sqrt{1553} by 8.

$$x=\frac{\sqrt{1553}-39}{4}$$

Now solve the equation x=\frac{-78±2\sqrt{1553}}{8} when ± is minus. Subtract 2\sqrt{1553} from -78.

$$x=\frac{-2\sqrt{1553}-78}{8}$$

Divide -78-2\sqrt{1553} by 8.

$$x=\frac{-\sqrt{1553}-39}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{1553}-39}{4} x=\frac{-\sqrt{1553}-39}{4}$$

Steps for Completing the Square

Subtract 4x from both sides.

$$4x^{2}+82x-4x=8$$

Combine 82x and -4x to get 78x.

$$4x^{2}+78x=8$$

Divide both sides by 4.

$$\frac{4x^{2}+78x}{4}=\frac{8}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{78}{4}x=\frac{8}{4}$$

Reduce the fraction \frac{78}{4} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{39}{2}x=\frac{8}{4}$$

Divide 8 by 4.

$$x^{2}+\frac{39}{2}x=2$$

Divide \frac{39}{2}, the coefficient of the x term, by 2 to get \frac{39}{4}. Then add the square of \frac{39}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{39}{2}x+\left(\frac{39}{4}\right)^{2}=2+\left(\frac{39}{4}\right)^{2}$$

Square \frac{39}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{39}{2}x+\frac{1521}{16}=2+\frac{1521}{16}$$

Add 2 to \frac{1521}{16}.

$$x^{2}+\frac{39}{2}x+\frac{1521}{16}=\frac{1553}{16}$$

Factor x^{2}+\frac{39}{2}x+\frac{1521}{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}.

$$\left(x+\frac{39}{4}\right)^{2}=\frac{1553}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{39}{4}\right)^{2}}=\sqrt{\frac{1553}{16}}$$

Simplify.

$$x+\frac{39}{4}=\frac{\sqrt{1553}}{4} x+\frac{39}{4}=-\frac{\sqrt{1553}}{4}$$

Subtract \frac{39}{4} from both sides of the equation.

$$x=\frac{\sqrt{1553}-39}{4} x=\frac{-\sqrt{1553}-39}{4}$$