Solve 3x+x^2=sqrt{45^2}

$$3x+ { x }^{ 2 } = \sqrt{ { 45 }^{ 2 } }$$

Solve for x

$$x = \frac{3 \sqrt{21} - 3}{2} \approx 5.373863542$$

$$x=\frac{-3\sqrt{21}-3}{2}\approx -8.373863542$$

Steps Using the Quadratic Formula

Calculate 45 to the power of 2 and get 2025.

$$3x+x^{2}=\sqrt{2025}$$

Calculate the square root of 2025 and get 45.

$$3x+x^{2}=45$$

Subtract 45 from both sides.

$$3x+x^{2}-45=0$$

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.

$$x^{2}+3x-45=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 3 for b, and -45 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-45\right)}}{2}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-45\right)}}{2}$$

Multiply -4 times -45.

$$x=\frac{-3±\sqrt{9+180}}{2}$$

Add 9 to 180.

$$x=\frac{-3±\sqrt{189}}{2}$$

Take the square root of 189.

$$x=\frac{-3±3\sqrt{21}}{2}$$

Now solve the equation x=\frac{-3±3\sqrt{21}}{2} when ± is plus. Add -3 to 3\sqrt{21}.

$$x=\frac{3\sqrt{21}-3}{2}$$

Now solve the equation x=\frac{-3±3\sqrt{21}}{2} when ± is minus. Subtract 3\sqrt{21} from -3.

$$x=\frac{-3\sqrt{21}-3}{2}$$

The equation is now solved.

$$x=\frac{3\sqrt{21}-3}{2} x=\frac{-3\sqrt{21}-3}{2}$$

Steps for Completing the Square

Calculate 45 to the power of 2 and get 2025.

$$3x+x^{2}=\sqrt{2025}$$

Calculate the square root of 2025 and get 45.

$$3x+x^{2}=45$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$x^{2}+3x=45$$

Divide 3, the coefficient of the x term, by 2 to get \frac{3}{2}. Then add the square of \frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=45+\left(\frac{3}{2}\right)^{2}$$

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=45+\frac{9}{4}$$

Add 45 to \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=\frac{189}{4}$$

Factor x^{2}+3x+\frac{9}{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}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{189}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{189}{4}}$$

Simplify.

$$x+\frac{3}{2}=\frac{3\sqrt{21}}{2} x+\frac{3}{2}=-\frac{3\sqrt{21}}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$x=\frac{3\sqrt{21}-3}{2} x=\frac{-3\sqrt{21}-3}{2}$$