Solve x^2+20x+17=-3

$${ x }^{ 2 } +20x+17=-3$$

Solve for x

$$x=4\sqrt{5}-10\approx -1.05572809$$

$$x=-4\sqrt{5}-10\approx -18.94427191$$

Steps Using the Quadratic Formula

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.

$$x^{2}+20x+17=-3$$

Add 3 to both sides of the equation.

$$x^{2}+20x+17-\left(-3\right)=-3-\left(-3\right)$$

Subtracting -3 from itself leaves 0.

$$x^{2}+20x+17-\left(-3\right)=0$$

Subtract -3 from 17.

$$x^{2}+20x+20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 20 for b, and 20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-20±\sqrt{20^{2}-4\times 20}}{2}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\times 20}}{2}$$

Multiply -4 times 20.

$$x=\frac{-20±\sqrt{400-80}}{2}$$

Add 400 to -80.

$$x=\frac{-20±\sqrt{320}}{2}$$

Take the square root of 320.

$$x=\frac{-20±8\sqrt{5}}{2}$$

Now solve the equation x=\frac{-20±8\sqrt{5}}{2} when ± is plus. Add -20 to 8\sqrt{5}.

$$x=\frac{8\sqrt{5}-20}{2}$$

Divide -20+8\sqrt{5} by 2.

$$x=4\sqrt{5}-10$$

Now solve the equation x=\frac{-20±8\sqrt{5}}{2} when ± is minus. Subtract 8\sqrt{5} from -20.

$$x=\frac{-8\sqrt{5}-20}{2}$$

Divide -20-8\sqrt{5} by 2.

$$x=-4\sqrt{5}-10$$

The equation is now solved.

$$x=4\sqrt{5}-10 x=-4\sqrt{5}-10$$

Steps for Completing the Square

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.

$$x^{2}+20x+17=-3$$

Subtract 17 from both sides of the equation.

$$x^{2}+20x+17-17=-3-17$$

Subtracting 17 from itself leaves 0.

$$x^{2}+20x=-3-17$$

Subtract 17 from -3.

$$x^{2}+20x=-20$$

Divide 20, the coefficient of the x term, by 2 to get 10. Then add the square of 10 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+20x+10^{2}=-20+10^{2}$$

Square 10.

$$x^{2}+20x+100=-20+100$$

Add -20 to 100.

$$x^{2}+20x+100=80$$

Factor x^{2}+20x+100. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+10\right)^{2}=80$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+10\right)^{2}}=\sqrt{80}$$

Simplify.

$$x+10=4\sqrt{5} x+10=-4\sqrt{5}$$

Subtract 10 from both sides of the equation.

$$x=4\sqrt{5}-10 x=-4\sqrt{5}-10$$