Solve 594=161l+20l^2

$$594 = 161 l + 20 l ^ { 2 }$$

Solve for l

$$l = \frac{11}{4} = 2\frac{3}{4} = 2.75$$

$$l = -\frac{54}{5} = -10\frac{4}{5} = -10.8$$

Steps Using Factoring By Grouping

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

$$161l+20l^{2}=594$$

Subtract 594 from both sides.

$$161l+20l^{2}-594=0$$

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

$$20l^{2}+161l-594=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 20l^{2}+al+bl-594. To find a and b, set up a system to be solved.

$$a+b=161 ab=20\left(-594\right)=-11880$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -11880.

$$-1,11880 -2,5940 -3,3960 -4,2970 -5,2376 -6,1980 -8,1485 -9,1320 -10,1188 -11,1080 -12,990 -15,792 -18,660 -20,594 -22,540 -24,495 -27,440 -30,396 -33,360 -36,330 -40,297 -44,270 -45,264 -54,220 -55,216 -60,198 -66,180 -72,165 -88,135 -90,132 -99,120 -108,110$$

Calculate the sum for each pair.

$$-1+11880=11879 -2+5940=5938 -3+3960=3957 -4+2970=2966 -5+2376=2371 -6+1980=1974 -8+1485=1477 -9+1320=1311 -10+1188=1178 -11+1080=1069 -12+990=978 -15+792=777 -18+660=642 -20+594=574 -22+540=518 -24+495=471 -27+440=413 -30+396=366 -33+360=327 -36+330=294 -40+297=257 -44+270=226 -45+264=219 -54+220=166 -55+216=161 -60+198=138 -66+180=114 -72+165=93 -88+135=47 -90+132=42 -99+120=21 -108+110=2$$

The solution is the pair that gives sum 161.

$$a=-55 b=216$$

Rewrite 20l^{2}+161l-594 as \left(20l^{2}-55l\right)+\left(216l-594\right).

$$\left(20l^{2}-55l\right)+\left(216l-594\right)$$

Factor out 5l in the first and 54 in the second group.

$$5l\left(4l-11\right)+54\left(4l-11\right)$$

Factor out common term 4l-11 by using distributive property.

$$\left(4l-11\right)\left(5l+54\right)$$

To find equation solutions, solve 4l-11=0 and 5l+54=0.

$$l=\frac{11}{4} l=-\frac{54}{5}$$

Steps Using the Quadratic Formula

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

$$161l+20l^{2}=594$$

Subtract 594 from both sides.

$$161l+20l^{2}-594=0$$

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.

$$20l^{2}+161l-594=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 20 for a, 161 for b, and -594 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$l=\frac{-161±\sqrt{161^{2}-4\times 20\left(-594\right)}}{2\times 20}$$

Square 161.

$$l=\frac{-161±\sqrt{25921-4\times 20\left(-594\right)}}{2\times 20}$$

Multiply -4 times 20.

$$l=\frac{-161±\sqrt{25921-80\left(-594\right)}}{2\times 20}$$

Multiply -80 times -594.

$$l=\frac{-161±\sqrt{25921+47520}}{2\times 20}$$

Add 25921 to 47520.

$$l=\frac{-161±\sqrt{73441}}{2\times 20}$$

Take the square root of 73441.

$$l=\frac{-161±271}{2\times 20}$$

Multiply 2 times 20.

$$l=\frac{-161±271}{40}$$

Now solve the equation l=\frac{-161±271}{40} when ± is plus. Add -161 to 271.

$$l=\frac{110}{40}$$

Reduce the fraction \frac{110}{40} to lowest terms by extracting and canceling out 10.

$$l=\frac{11}{4}$$

Now solve the equation l=\frac{-161±271}{40} when ± is minus. Subtract 271 from -161.

$$l=-\frac{432}{40}$$

Reduce the fraction \frac{-432}{40} to lowest terms by extracting and canceling out 8.

$$l=-\frac{54}{5}$$

The equation is now solved.

$$l=\frac{11}{4} l=-\frac{54}{5}$$

Steps for Completing the Square

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

$$161l+20l^{2}=594$$

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.

$$20l^{2}+161l=594$$

Divide both sides by 20.

$$\frac{20l^{2}+161l}{20}=\frac{594}{20}$$

Dividing by 20 undoes the multiplication by 20.

$$l^{2}+\frac{161}{20}l=\frac{594}{20}$$

Reduce the fraction \frac{594}{20} to lowest terms by extracting and canceling out 2.

$$l^{2}+\frac{161}{20}l=\frac{297}{10}$$

Divide \frac{161}{20}, the coefficient of the x term, by 2 to get \frac{161}{40}. Then add the square of \frac{161}{40} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$l^{2}+\frac{161}{20}l+\left(\frac{161}{40}\right)^{2}=\frac{297}{10}+\left(\frac{161}{40}\right)^{2}$$

Square \frac{161}{40} by squaring both the numerator and the denominator of the fraction.

$$l^{2}+\frac{161}{20}l+\frac{25921}{1600}=\frac{297}{10}+\frac{25921}{1600}$$

Add \frac{297}{10} to \frac{25921}{1600} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$l^{2}+\frac{161}{20}l+\frac{25921}{1600}=\frac{73441}{1600}$$

Factor l^{2}+\frac{161}{20}l+\frac{25921}{1600}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(l+\frac{161}{40}\right)^{2}=\frac{73441}{1600}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(l+\frac{161}{40}\right)^{2}}=\sqrt{\frac{73441}{1600}}$$

Simplify.

$$l+\frac{161}{40}=\frac{271}{40} l+\frac{161}{40}=-\frac{271}{40}$$

Subtract \frac{161}{40} from both sides of the equation.

$$l=\frac{11}{4} l=-\frac{54}{5}$$