Solve 0=-1/2*4.91*x^2-0.1x+5.5

$$0=- \frac{ 1 }{ 2 } \times 4.91 \times { x }^{ 2 } -0.1x+5.5$$

Solve for x

$$x = \frac{10 \sqrt{5402} - 10}{491} \approx 1.476543774$$

$$x=\frac{-10\sqrt{5402}-10}{491}\approx -1.517276971$$

Steps Using the Quadratic Formula

Multiply -\frac{1}{2} and 4.91 to get -\frac{491}{200}.

$$0=-\frac{491}{200}x^{2}-0.1x+5.5$$

Swap sides so that all variable terms are on the left hand side.

$$-\frac{491}{200}x^{2}-0.1x+5.5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{491}{200} for a, -0.1 for b, and 5.5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-0.1\right)±\sqrt{\left(-0.1\right)^{2}-4\left(-\frac{491}{200}\right)\times 5.5}}{2\left(-\frac{491}{200}\right)}$$

Square -0.1 by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-0.1\right)±\sqrt{0.01-4\left(-\frac{491}{200}\right)\times 5.5}}{2\left(-\frac{491}{200}\right)}$$

Multiply -4 times -\frac{491}{200}.

$$x=\frac{-\left(-0.1\right)±\sqrt{0.01+\frac{491}{50}\times 5.5}}{2\left(-\frac{491}{200}\right)}$$

Multiply \frac{491}{50} times 5.5 by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-0.1\right)±\sqrt{\frac{1+5401}{100}}}{2\left(-\frac{491}{200}\right)}$$

Add 0.01 to \frac{5401}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-0.1\right)±\sqrt{\frac{2701}{50}}}{2\left(-\frac{491}{200}\right)}$$

Take the square root of \frac{2701}{50}.

$$x=\frac{-\left(-0.1\right)±\frac{\sqrt{5402}}{10}}{2\left(-\frac{491}{200}\right)}$$

The opposite of -0.1 is 0.1.

$$x=\frac{0.1±\frac{\sqrt{5402}}{10}}{2\left(-\frac{491}{200}\right)}$$

Multiply 2 times -\frac{491}{200}.

$$x=\frac{0.1±\frac{\sqrt{5402}}{10}}{-\frac{491}{100}}$$

Now solve the equation x=\frac{0.1±\frac{\sqrt{5402}}{10}}{-\frac{491}{100}} when ± is plus. Add 0.1 to \frac{\sqrt{5402}}{10}.

$$x=\frac{\sqrt{5402}+1}{-\frac{491}{100}\times 10}$$

Divide \frac{1+\sqrt{5402}}{10} by -\frac{491}{100} by multiplying \frac{1+\sqrt{5402}}{10} by the reciprocal of -\frac{491}{100}.

$$x=\frac{-10\sqrt{5402}-10}{491}$$

Now solve the equation x=\frac{0.1±\frac{\sqrt{5402}}{10}}{-\frac{491}{100}} when ± is minus. Subtract \frac{\sqrt{5402}}{10} from 0.1.

$$x=\frac{1-\sqrt{5402}}{-\frac{491}{100}\times 10}$$

Divide \frac{1-\sqrt{5402}}{10} by -\frac{491}{100} by multiplying \frac{1-\sqrt{5402}}{10} by the reciprocal of -\frac{491}{100}.

$$x=\frac{10\sqrt{5402}-10}{491}$$

The equation is now solved.

$$x=\frac{-10\sqrt{5402}-10}{491} x=\frac{10\sqrt{5402}-10}{491}$$

Steps for Completing the Square

Multiply -\frac{1}{2} and 4.91 to get -\frac{491}{200}.

$$0=-\frac{491}{200}x^{2}-0.1x+5.5$$

Swap sides so that all variable terms are on the left hand side.

$$-\frac{491}{200}x^{2}-0.1x+5.5=0$$

Subtract 5.5 from both sides. Anything subtracted from zero gives its negation.

$$-\frac{491}{200}x^{2}-0.1x=-5.5$$

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.

$$-\frac{491}{200}x^{2}-0.1x=-\frac{11}{2}$$

Divide both sides of the equation by -\frac{491}{200}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{491}{200}x^{2}-0.1x}{-\frac{491}{200}}=-\frac{\frac{11}{2}}{-\frac{491}{200}}$$

Dividing by -\frac{491}{200} undoes the multiplication by -\frac{491}{200}.

$$x^{2}+\left(-\frac{0.1}{-\frac{491}{200}}\right)x=-\frac{\frac{11}{2}}{-\frac{491}{200}}$$

Divide -0.1 by -\frac{491}{200} by multiplying -0.1 by the reciprocal of -\frac{491}{200}.

$$x^{2}+\frac{20}{491}x=-\frac{\frac{11}{2}}{-\frac{491}{200}}$$

Divide -\frac{11}{2} by -\frac{491}{200} by multiplying -\frac{11}{2} by the reciprocal of -\frac{491}{200}.

$$x^{2}+\frac{20}{491}x=\frac{1100}{491}$$

Divide \frac{20}{491}, the coefficient of the x term, by 2 to get \frac{10}{491}. Then add the square of \frac{10}{491} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{20}{491}x+\left(\frac{10}{491}\right)^{2}=\frac{1100}{491}+\left(\frac{10}{491}\right)^{2}$$

Square \frac{10}{491} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{20}{491}x+\frac{100}{241081}=\frac{1100}{491}+\frac{100}{241081}$$

Add \frac{1100}{491} to \frac{100}{241081} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{20}{491}x+\frac{100}{241081}=\frac{540200}{241081}$$

Factor x^{2}+\frac{20}{491}x+\frac{100}{241081}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{10}{491}\right)^{2}=\frac{540200}{241081}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{10}{491}\right)^{2}}=\sqrt{\frac{540200}{241081}}$$

Simplify.

$$x+\frac{10}{491}=\frac{10\sqrt{5402}}{491} x+\frac{10}{491}=-\frac{10\sqrt{5402}}{491}$$

Subtract \frac{10}{491} from both sides of the equation.

$$x=\frac{10\sqrt{5402}-10}{491} x=\frac{-10\sqrt{5402}-10}{491}$$