Solve 2019x^2=2020+[x]

$$2019 x ^ { 2 } = 2020 + [ x ]$$

Solve for x

$$x=-1$$

$$x = \frac{2020}{2019} = 1\frac{1}{2019} \approx 1.000495295$$

Steps Using Factoring By Grouping

Subtract 2020 from both sides.

$$2019x^{2}-2020=x$$

Subtract x from both sides.

$$2019x^{2}-2020-x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$2019x^{2}-x-2020=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2019x^{2}+ax+bx-2020. To find a and b, set up a system to be solved.

$$a+b=-1 ab=2019\left(-2020\right)=-4078380$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -4078380.

$$1,-4078380 2,-2039190 3,-1359460 4,-1019595 5,-815676 6,-679730 10,-407838 12,-339865 15,-271892 20,-203919 30,-135946 60,-67973 101,-40380 202,-20190 303,-13460 404,-10095 505,-8076 606,-6730 673,-6060 1010,-4038 1212,-3365 1346,-3030 1515,-2692 2019,-2020$$

Calculate the sum for each pair.

$$1-4078380=-4078379 2-2039190=-2039188 3-1359460=-1359457 4-1019595=-1019591 5-815676=-815671 6-679730=-679724 10-407838=-407828 12-339865=-339853 15-271892=-271877 20-203919=-203899 30-135946=-135916 60-67973=-67913 101-40380=-40279 202-20190=-19988 303-13460=-13157 404-10095=-9691 505-8076=-7571 606-6730=-6124 673-6060=-5387 1010-4038=-3028 1212-3365=-2153 1346-3030=-1684 1515-2692=-1177 2019-2020=-1$$

The solution is the pair that gives sum -1.

$$a=-2020 b=2019$$

Rewrite 2019x^{2}-x-2020 as \left(2019x^{2}-2020x\right)+\left(2019x-2020\right).

$$\left(2019x^{2}-2020x\right)+\left(2019x-2020\right)$$

Factor out x in 2019x^{2}-2020x.

$$x\left(2019x-2020\right)+2019x-2020$$

Factor out common term 2019x-2020 by using distributive property.

$$\left(2019x-2020\right)\left(x+1\right)$$

To find equation solutions, solve 2019x-2020=0 and x+1=0.

$$x=\frac{2020}{2019} x=-1$$

Steps Using the Quadratic Formula

Subtract 2020 from both sides.

$$2019x^{2}-2020=x$$

Subtract x from both sides.

$$2019x^{2}-2020-x=0$$

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.

$$2019x^{2}-x-2020=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2019 for a, -1 for b, and -2020 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 2019\left(-2020\right)}}{2\times 2019}$$

Multiply -4 times 2019.

$$x=\frac{-\left(-1\right)±\sqrt{1-8076\left(-2020\right)}}{2\times 2019}$$

Multiply -8076 times -2020.

$$x=\frac{-\left(-1\right)±\sqrt{1+16313520}}{2\times 2019}$$

Add 1 to 16313520.

$$x=\frac{-\left(-1\right)±\sqrt{16313521}}{2\times 2019}$$

Take the square root of 16313521.

$$x=\frac{-\left(-1\right)±4039}{2\times 2019}$$

The opposite of -1 is 1.

$$x=\frac{1±4039}{2\times 2019}$$

Multiply 2 times 2019.

$$x=\frac{1±4039}{4038}$$

Now solve the equation x=\frac{1±4039}{4038} when ± is plus. Add 1 to 4039.

$$x=\frac{4040}{4038}$$

Reduce the fraction \frac{4040}{4038} to lowest terms by extracting and canceling out 2.

$$x=\frac{2020}{2019}$$

Now solve the equation x=\frac{1±4039}{4038} when ± is minus. Subtract 4039 from 1.

$$x=-\frac{4038}{4038}$$

Divide -4038 by 4038.

$$x=-1$$

The equation is now solved.

$$x=\frac{2020}{2019} x=-1$$

Steps for Completing the Square

Subtract x from both sides.

$$2019x^{2}-x=2020$$

Divide both sides by 2019.

$$\frac{2019x^{2}-x}{2019}=\frac{2020}{2019}$$

Dividing by 2019 undoes the multiplication by 2019.

$$x^{2}-\frac{1}{2019}x=\frac{2020}{2019}$$

Divide -\frac{1}{2019}, the coefficient of the x term, by 2 to get -\frac{1}{4038}. Then add the square of -\frac{1}{4038} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{2019}x+\left(-\frac{1}{4038}\right)^{2}=\frac{2020}{2019}+\left(-\frac{1}{4038}\right)^{2}$$

Square -\frac{1}{4038} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{2019}x+\frac{1}{16305444}=\frac{2020}{2019}+\frac{1}{16305444}$$

Add \frac{2020}{2019} to \frac{1}{16305444} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{2019}x+\frac{1}{16305444}=\frac{16313521}{16305444}$$

Factor x^{2}-\frac{1}{2019}x+\frac{1}{16305444}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{4038}\right)^{2}=\frac{16313521}{16305444}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4038}\right)^{2}}=\sqrt{\frac{16313521}{16305444}}$$

Simplify.

$$x-\frac{1}{4038}=\frac{4039}{4038} x-\frac{1}{4038}=-\frac{4039}{4038}$$

Add \frac{1}{4038} to both sides of the equation.

$$x=\frac{2020}{2019} x=-1$$