Solve 4x^2+42x=2x-8

$$4 x ^ { 2 } + 42 x = 2 x - 8$$

Solve for x (complex solution)

$$x=\sqrt{23}-5\approx -0.204168477$$

$$x=-\left(\sqrt{23}+5\right)\approx -9.795831523$$

Steps Using the Quadratic Formula

Subtract 2x from both sides.

$$4x^{2}+42x-2x=-8$$

Combine 42x and -2x to get 40x.

$$4x^{2}+40x=-8$$

Add 8 to both sides.

$$4x^{2}+40x+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 40 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-40±\sqrt{40^{2}-4\times 4\times 8}}{2\times 4}$$

Square 40.

$$x=\frac{-40±\sqrt{1600-4\times 4\times 8}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-40±\sqrt{1600-16\times 8}}{2\times 4}$$

Multiply -16 times 8.

$$x=\frac{-40±\sqrt{1600-128}}{2\times 4}$$

Add 1600 to -128.

$$x=\frac{-40±\sqrt{1472}}{2\times 4}$$

Take the square root of 1472.

$$x=\frac{-40±8\sqrt{23}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-40±8\sqrt{23}}{8}$$

Now solve the equation x=\frac{-40±8\sqrt{23}}{8} when ± is plus. Add -40 to 8\sqrt{23}.

$$x=\frac{8\sqrt{23}-40}{8}$$

Divide -40+8\sqrt{23} by 8.

$$x=\sqrt{23}-5$$

Now solve the equation x=\frac{-40±8\sqrt{23}}{8} when ± is minus. Subtract 8\sqrt{23} from -40.

$$x=\frac{-8\sqrt{23}-40}{8}$$

Divide -40-8\sqrt{23} by 8.

$$x=-\sqrt{23}-5$$

The equation is now solved.

$$x=\sqrt{23}-5 x=-\sqrt{23}-5$$

Steps for Completing the Square

Subtract 2x from both sides.

$$4x^{2}+42x-2x=-8$$

Combine 42x and -2x to get 40x.

$$4x^{2}+40x=-8$$

Divide both sides by 4.

$$\frac{4x^{2}+40x}{4}=-\frac{8}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{40}{4}x=-\frac{8}{4}$$

Divide 40 by 4.

$$x^{2}+10x=-\frac{8}{4}$$

Divide -8 by 4.

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

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.

$$x^{2}+10x+5^{2}=-2+5^{2}$$

Square 5.

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

Add -2 to 25.

$$x^{2}+10x+25=23$$

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}.

$$\left(x+5\right)^{2}=23$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{23}$$

Simplify.

$$x+5=\sqrt{23} x+5=-\sqrt{23}$$

Subtract 5 from both sides of the equation.

$$x=\sqrt{23}-5 x=-\sqrt{23}-5$$

Solve for x

$$x=\sqrt{23}-5\approx -0.204168477$$

$$x=-\sqrt{23}-5\approx -9.795831523$$

Steps Using the Quadratic Formula

Subtract 2x from both sides.

$$4x^{2}+42x-2x=-8$$

Combine 42x and -2x to get 40x.

$$4x^{2}+40x=-8$$

Add 8 to both sides.

$$4x^{2}+40x+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 40 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-40±\sqrt{40^{2}-4\times 4\times 8}}{2\times 4}$$

Square 40.

$$x=\frac{-40±\sqrt{1600-4\times 4\times 8}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-40±\sqrt{1600-16\times 8}}{2\times 4}$$

Multiply -16 times 8.

$$x=\frac{-40±\sqrt{1600-128}}{2\times 4}$$

Add 1600 to -128.

$$x=\frac{-40±\sqrt{1472}}{2\times 4}$$

Take the square root of 1472.

$$x=\frac{-40±8\sqrt{23}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-40±8\sqrt{23}}{8}$$

Now solve the equation x=\frac{-40±8\sqrt{23}}{8} when ± is plus. Add -40 to 8\sqrt{23}.

$$x=\frac{8\sqrt{23}-40}{8}$$

Divide -40+8\sqrt{23} by 8.

$$x=\sqrt{23}-5$$

Now solve the equation x=\frac{-40±8\sqrt{23}}{8} when ± is minus. Subtract 8\sqrt{23} from -40.

$$x=\frac{-8\sqrt{23}-40}{8}$$

Divide -40-8\sqrt{23} by 8.

$$x=-\sqrt{23}-5$$

The equation is now solved.

$$x=\sqrt{23}-5 x=-\sqrt{23}-5$$

Steps for Completing the Square

Subtract 2x from both sides.

$$4x^{2}+42x-2x=-8$$

Combine 42x and -2x to get 40x.

$$4x^{2}+40x=-8$$

Divide both sides by 4.

$$\frac{4x^{2}+40x}{4}=-\frac{8}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{40}{4}x=-\frac{8}{4}$$

Divide 40 by 4.

$$x^{2}+10x=-\frac{8}{4}$$

Divide -8 by 4.

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

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.

$$x^{2}+10x+5^{2}=-2+5^{2}$$

Square 5.

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

Add -2 to 25.

$$x^{2}+10x+25=23$$

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}.

$$\left(x+5\right)^{2}=23$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{23}$$

Simplify.

$$x+5=\sqrt{23} x+5=-\sqrt{23}$$

Subtract 5 from both sides of the equation.

$$x=\sqrt{23}-5 x=-\sqrt{23}-5$$