Solve x^2+2x+3x=1

$$x ^ { 2 } + 2 x + 3 x = 1$$

Solve for x

$$x=\frac{\sqrt{29}-5}{2}\approx 0.192582404$$

$$x=\frac{-\sqrt{29}-5}{2}\approx -5.192582404$$

Steps Using the Quadratic Formula

Combine 2x and 3x to get 5x.

$$x^{2}+5x=1$$

Subtract 1 from both sides.

$$x^{2}+5x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 5 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-1\right)}}{2}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-1\right)}}{2}$$

Multiply -4 times -1.

$$x=\frac{-5±\sqrt{25+4}}{2}$$

Add 25 to 4.

$$x=\frac{-5±\sqrt{29}}{2}$$

Now solve the equation x=\frac{-5±\sqrt{29}}{2} when ± is plus. Add -5 to \sqrt{29}.

$$x=\frac{\sqrt{29}-5}{2}$$

Now solve the equation x=\frac{-5±\sqrt{29}}{2} when ± is minus. Subtract \sqrt{29} from -5.

$$x=\frac{-\sqrt{29}-5}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{29}-5}{2} x=\frac{-\sqrt{29}-5}{2}$$

Steps for Completing the Square

Combine 2x and 3x to get 5x.

$$x^{2}+5x=1$$

Divide 5, the coefficient of the x term, by 2 to get \frac{5}{2}. Then add the square of \frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=1+\left(\frac{5}{2}\right)^{2}$$

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+5x+\frac{25}{4}=1+\frac{25}{4}$$

Add 1 to \frac{25}{4}.

$$x^{2}+5x+\frac{25}{4}=\frac{29}{4}$$

Factor x^{2}+5x+\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{29}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{29}{4}}$$

Simplify.

$$x+\frac{5}{2}=\frac{\sqrt{29}}{2} x+\frac{5}{2}=-\frac{\sqrt{29}}{2}$$

Subtract \frac{5}{2} from both sides of the equation.

$$x=\frac{\sqrt{29}-5}{2} x=\frac{-\sqrt{29}-5}{2}$$