Solve 1x^2+200x=50

$$1 x ^ { 2 } + 200 x = 50$$

Solve for x

$$x=5\sqrt{402}-100\approx 0.249688279$$

$$x=-5\sqrt{402}-100\approx -200.249688279$$

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.

$$x^{2}+200x=50$$

Subtract 50 from both sides of the equation.

$$x^{2}+200x-50=50-50$$

Subtracting 50 from itself leaves 0.

$$x^{2}+200x-50=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 200 for b, and -50 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-200±\sqrt{200^{2}-4\left(-50\right)}}{2}$$

Square 200.

$$x=\frac{-200±\sqrt{40000-4\left(-50\right)}}{2}$$

Multiply -4 times -50.

$$x=\frac{-200±\sqrt{40000+200}}{2}$$

Add 40000 to 200.

$$x=\frac{-200±\sqrt{40200}}{2}$$

Take the square root of 40200.

$$x=\frac{-200±10\sqrt{402}}{2}$$

Now solve the equation x=\frac{-200±10\sqrt{402}}{2} when ± is plus. Add -200 to 10\sqrt{402}.

$$x=\frac{10\sqrt{402}-200}{2}$$

Divide -200+10\sqrt{402} by 2.

$$x=5\sqrt{402}-100$$

Now solve the equation x=\frac{-200±10\sqrt{402}}{2} when ± is minus. Subtract 10\sqrt{402} from -200.

$$x=\frac{-10\sqrt{402}-200}{2}$$

Divide -200-10\sqrt{402} by 2.

$$x=-5\sqrt{402}-100$$

The equation is now solved.

$$x=5\sqrt{402}-100 x=-5\sqrt{402}-100$$

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.

$$x^{2}+200x=50$$

Divide 200, the coefficient of the x term, by 2 to get 100. Then add the square of 100 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+200x+100^{2}=50+100^{2}$$

Square 100.

$$x^{2}+200x+10000=50+10000$$

Add 50 to 10000.

$$x^{2}+200x+10000=10050$$

Factor x^{2}+200x+10000. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+100\right)^{2}=10050$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+100\right)^{2}}=\sqrt{10050}$$

Simplify.

$$x+100=5\sqrt{402} x+100=-5\sqrt{402}$$

Subtract 100 from both sides of the equation.

$$x=5\sqrt{402}-100 x=-5\sqrt{402}-100$$