Solve -2(x+5)^2-1=49

$$- 2 ( x + 5 ) ^ { 2 } - 1 = 49$$

Solve for x (complex solution)

$$x=-5-5i$$

$$x=-5+5i$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+5\right)^{2}.

$$-2\left(x^{2}+10x+25\right)-1=49$$

Use the distributive property to multiply -2 by x^{2}+10x+25.

$$-2x^{2}-20x-50-1=49$$

Subtract 1 from -50 to get -51.

$$-2x^{2}-20x-51=49$$

Subtract 49 from both sides.

$$-2x^{2}-20x-51-49=0$$

Subtract 49 from -51 to get -100.

$$-2x^{2}-20x-100=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -20 for b, and -100 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\left(-2\right)\left(-100\right)}}{2\left(-2\right)}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\left(-2\right)\left(-100\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\left(-20\right)±\sqrt{400+8\left(-100\right)}}{2\left(-2\right)}$$

Multiply 8 times -100.

$$x=\frac{-\left(-20\right)±\sqrt{400-800}}{2\left(-2\right)}$$

Add 400 to -800.

$$x=\frac{-\left(-20\right)±\sqrt{-400}}{2\left(-2\right)}$$

Take the square root of -400.

$$x=\frac{-\left(-20\right)±20i}{2\left(-2\right)}$$

The opposite of -20 is 20.

$$x=\frac{20±20i}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{20±20i}{-4}$$

Now solve the equation x=\frac{20±20i}{-4} when ± is plus. Add 20 to 20i.

$$x=\frac{20+20i}{-4}$$

Divide 20+20i by -4.

$$x=-5-5i$$

Now solve the equation x=\frac{20±20i}{-4} when ± is minus. Subtract 20i from 20.

$$x=\frac{20-20i}{-4}$$

Divide 20-20i by -4.

$$x=-5+5i$$

The equation is now solved.

$$x=-5-5i x=-5+5i$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+5\right)^{2}.

$$-2\left(x^{2}+10x+25\right)-1=49$$

Use the distributive property to multiply -2 by x^{2}+10x+25.

$$-2x^{2}-20x-50-1=49$$

Subtract 1 from -50 to get -51.

$$-2x^{2}-20x-51=49$$

Add 51 to both sides.

$$-2x^{2}-20x=49+51$$

Add 49 and 51 to get 100.

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

Divide both sides by -2.

$$\frac{-2x^{2}-20x}{-2}=\frac{100}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\left(-\frac{20}{-2}\right)x=\frac{100}{-2}$$

Divide -20 by -2.

$$x^{2}+10x=\frac{100}{-2}$$

Divide 100 by -2.

$$x^{2}+10x=-50$$

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.

$$x^{2}+10x+5^{2}=-50+5^{2}$$

Square 5.

$$x^{2}+10x+25=-50+25$$

Add -50 to 25.

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

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}.

$$\left(x+5\right)^{2}=-25$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{-25}$$

Simplify.

$$x+5=5i x+5=-5i$$

Subtract 5 from both sides of the equation.

$$x=-5+5i x=-5-5i$$