Solve 60x^2+57x=18

$$60 x ^ { 2 } + 57 x = 18$$

Solve for x

$$x = -\frac{6}{5} = -1\frac{1}{5} = -1.2$$

$$x=\frac{1}{4}=0.25$$

Steps Using Factoring By Grouping

Subtract 18 from both sides.

$$60x^{2}+57x-18=0$$

Divide both sides by 3.

$$20x^{2}+19x-6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 20x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=19 ab=20\left(-6\right)=-120$$

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 -120.

$$-1,120 -2,60 -3,40 -4,30 -5,24 -6,20 -8,15 -10,12$$

Calculate the sum for each pair.

$$-1+120=119 -2+60=58 -3+40=37 -4+30=26 -5+24=19 -6+20=14 -8+15=7 -10+12=2$$

The solution is the pair that gives sum 19.

$$a=-5 b=24$$

Rewrite 20x^{2}+19x-6 as \left(20x^{2}-5x\right)+\left(24x-6\right).

$$\left(20x^{2}-5x\right)+\left(24x-6\right)$$

Factor out 5x in the first and 6 in the second group.

$$5x\left(4x-1\right)+6\left(4x-1\right)$$

Factor out common term 4x-1 by using distributive property.

$$\left(4x-1\right)\left(5x+6\right)$$

To find equation solutions, solve 4x-1=0 and 5x+6=0.

$$x=\frac{1}{4} x=-\frac{6}{5}$$

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.

$$60x^{2}+57x=18$$

Subtract 18 from both sides of the equation.

$$60x^{2}+57x-18=18-18$$

Subtracting 18 from itself leaves 0.

$$60x^{2}+57x-18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 60 for a, 57 for b, and -18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-57±\sqrt{57^{2}-4\times 60\left(-18\right)}}{2\times 60}$$

Square 57.

$$x=\frac{-57±\sqrt{3249-4\times 60\left(-18\right)}}{2\times 60}$$

Multiply -4 times 60.

$$x=\frac{-57±\sqrt{3249-240\left(-18\right)}}{2\times 60}$$

Multiply -240 times -18.

$$x=\frac{-57±\sqrt{3249+4320}}{2\times 60}$$

Add 3249 to 4320.

$$x=\frac{-57±\sqrt{7569}}{2\times 60}$$

Take the square root of 7569.

$$x=\frac{-57±87}{2\times 60}$$

Multiply 2 times 60.

$$x=\frac{-57±87}{120}$$

Now solve the equation x=\frac{-57±87}{120} when ± is plus. Add -57 to 87.

$$x=\frac{30}{120}$$

Reduce the fraction \frac{30}{120} to lowest terms by extracting and canceling out 30.

$$x=\frac{1}{4}$$

Now solve the equation x=\frac{-57±87}{120} when ± is minus. Subtract 87 from -57.

$$x=-\frac{144}{120}$$

Reduce the fraction \frac{-144}{120} to lowest terms by extracting and canceling out 24.

$$x=-\frac{6}{5}$$

The equation is now solved.

$$x=\frac{1}{4} x=-\frac{6}{5}$$

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.

$$60x^{2}+57x=18$$

Divide both sides by 60.

$$\frac{60x^{2}+57x}{60}=\frac{18}{60}$$

Dividing by 60 undoes the multiplication by 60.

$$x^{2}+\frac{57}{60}x=\frac{18}{60}$$

Reduce the fraction \frac{57}{60} to lowest terms by extracting and canceling out 3.

$$x^{2}+\frac{19}{20}x=\frac{18}{60}$$

Reduce the fraction \frac{18}{60} to lowest terms by extracting and canceling out 6.

$$x^{2}+\frac{19}{20}x=\frac{3}{10}$$

Divide \frac{19}{20}, the coefficient of the x term, by 2 to get \frac{19}{40}. Then add the square of \frac{19}{40} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{19}{20}x+\left(\frac{19}{40}\right)^{2}=\frac{3}{10}+\left(\frac{19}{40}\right)^{2}$$

Square \frac{19}{40} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{19}{20}x+\frac{361}{1600}=\frac{3}{10}+\frac{361}{1600}$$

Add \frac{3}{10} to \frac{361}{1600} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{19}{20}x+\frac{361}{1600}=\frac{841}{1600}$$

Factor x^{2}+\frac{19}{20}x+\frac{361}{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}.

$$\left(x+\frac{19}{40}\right)^{2}=\frac{841}{1600}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{19}{40}\right)^{2}}=\sqrt{\frac{841}{1600}}$$

Simplify.

$$x+\frac{19}{40}=\frac{29}{40} x+\frac{19}{40}=-\frac{29}{40}$$

Subtract \frac{19}{40} from both sides of the equation.

$$x=\frac{1}{4} x=-\frac{6}{5}$$