Solve 0=x^2+10x+4

$$0 = x ^ { 2 } + 10 x + 4$$

Solve for x (complex solution)

$$x=\sqrt{21}-5\approx -0.417424305$$

$$x=-\left(\sqrt{21}+5\right)\approx -9.582575695$$

Steps Using the Quadratic Formula

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

$$x^{2}+10x+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 10 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\times 4}}{2}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\times 4}}{2}$$

Multiply -4 times 4.

$$x=\frac{-10±\sqrt{100-16}}{2}$$

Add 100 to -16.

$$x=\frac{-10±\sqrt{84}}{2}$$

Take the square root of 84.

$$x=\frac{-10±2\sqrt{21}}{2}$$

Now solve the equation x=\frac{-10±2\sqrt{21}}{2} when ± is plus. Add -10 to 2\sqrt{21}.

$$x=\frac{2\sqrt{21}-10}{2}$$

Divide -10+2\sqrt{21} by 2.

$$x=\sqrt{21}-5$$

Now solve the equation x=\frac{-10±2\sqrt{21}}{2} when ± is minus. Subtract 2\sqrt{21} from -10.

$$x=\frac{-2\sqrt{21}-10}{2}$$

Divide -10-2\sqrt{21} by 2.

$$x=-\sqrt{21}-5$$

The equation is now solved.

$$x=\sqrt{21}-5 x=-\sqrt{21}-5$$

Steps for Completing the Square

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

$$x^{2}+10x+4=0$$

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}+10x=-4$$

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.

$$x^{2}+10x+5^{2}=-4+5^{2}$$

Square 5.

$$x^{2}+10x+25=-4+25$$

Add -4 to 25.

$$x^{2}+10x+25=21$$

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}.

$$\left(x+5\right)^{2}=21$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{21}$$

Simplify.

$$x+5=\sqrt{21} x+5=-\sqrt{21}$$

Subtract 5 from both sides of the equation.

$$x=\sqrt{21}-5 x=-\sqrt{21}-5$$

Solve for x

$$x=\sqrt{21}-5\approx -0.417424305$$

$$x=-\sqrt{21}-5\approx -9.582575695$$

Steps Using the Quadratic Formula

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

$$x^{2}+10x+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 10 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\times 4}}{2}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\times 4}}{2}$$

Multiply -4 times 4.

$$x=\frac{-10±\sqrt{100-16}}{2}$$

Add 100 to -16.

$$x=\frac{-10±\sqrt{84}}{2}$$

Take the square root of 84.

$$x=\frac{-10±2\sqrt{21}}{2}$$

Now solve the equation x=\frac{-10±2\sqrt{21}}{2} when ± is plus. Add -10 to 2\sqrt{21}.

$$x=\frac{2\sqrt{21}-10}{2}$$

Divide -10+2\sqrt{21} by 2.

$$x=\sqrt{21}-5$$

Now solve the equation x=\frac{-10±2\sqrt{21}}{2} when ± is minus. Subtract 2\sqrt{21} from -10.

$$x=\frac{-2\sqrt{21}-10}{2}$$

Divide -10-2\sqrt{21} by 2.

$$x=-\sqrt{21}-5$$

The equation is now solved.

$$x=\sqrt{21}-5 x=-\sqrt{21}-5$$

Steps for Completing the Square

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

$$x^{2}+10x+4=0$$

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}+10x=-4$$

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.

$$x^{2}+10x+5^{2}=-4+5^{2}$$

Square 5.

$$x^{2}+10x+25=-4+25$$

Add -4 to 25.

$$x^{2}+10x+25=21$$

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}.

$$\left(x+5\right)^{2}=21$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{21}$$

Simplify.

$$x+5=\sqrt{21} x+5=-\sqrt{21}$$

Subtract 5 from both sides of the equation.

$$x=\sqrt{21}-5 x=-\sqrt{21}-5$$