Solve 54x^2-21x-68=0

$$54 { x }^{ 2 } -21x-68=0$$

Solve for x

$$x=-\frac{17}{18}\approx -0.944444444$$

$$x = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$$

Steps Using Factoring By Grouping

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 54x^{2}+ax+bx-68. To find a and b, set up a system to be solved.

$$a+b=-21 ab=54\left(-68\right)=-3672$$

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 -3672.

$$1,-3672 2,-1836 3,-1224 4,-918 6,-612 8,-459 9,-408 12,-306 17,-216 18,-204 24,-153 27,-136 34,-108 36,-102 51,-72 54,-68$$

Calculate the sum for each pair.

$$1-3672=-3671 2-1836=-1834 3-1224=-1221 4-918=-914 6-612=-606 8-459=-451 9-408=-399 12-306=-294 17-216=-199 18-204=-186 24-153=-129 27-136=-109 34-108=-74 36-102=-66 51-72=-21 54-68=-14$$

The solution is the pair that gives sum -21.

$$a=-72 b=51$$

Rewrite 54x^{2}-21x-68 as \left(54x^{2}-72x\right)+\left(51x-68\right).

$$\left(54x^{2}-72x\right)+\left(51x-68\right)$$

Factor out 18x in the first and 17 in the second group.

$$18x\left(3x-4\right)+17\left(3x-4\right)$$

Factor out common term 3x-4 by using distributive property.

$$\left(3x-4\right)\left(18x+17\right)$$

To find equation solutions, solve 3x-4=0 and 18x+17=0.

$$x=\frac{4}{3} x=-\frac{17}{18}$$

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.

$$54x^{2}-21x-68=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 54 for a, -21 for b, and -68 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-21\right)±\sqrt{\left(-21\right)^{2}-4\times 54\left(-68\right)}}{2\times 54}$$

Square -21.

$$x=\frac{-\left(-21\right)±\sqrt{441-4\times 54\left(-68\right)}}{2\times 54}$$

Multiply -4 times 54.

$$x=\frac{-\left(-21\right)±\sqrt{441-216\left(-68\right)}}{2\times 54}$$

Multiply -216 times -68.

$$x=\frac{-\left(-21\right)±\sqrt{441+14688}}{2\times 54}$$

Add 441 to 14688.

$$x=\frac{-\left(-21\right)±\sqrt{15129}}{2\times 54}$$

Take the square root of 15129.

$$x=\frac{-\left(-21\right)±123}{2\times 54}$$

The opposite of -21 is 21.

$$x=\frac{21±123}{2\times 54}$$

Multiply 2 times 54.

$$x=\frac{21±123}{108}$$

Now solve the equation x=\frac{21±123}{108} when ± is plus. Add 21 to 123.

$$x=\frac{144}{108}$$

Reduce the fraction \frac{144}{108} to lowest terms by extracting and canceling out 36.

$$x=\frac{4}{3}$$

Now solve the equation x=\frac{21±123}{108} when ± is minus. Subtract 123 from 21.

$$x=-\frac{102}{108}$$

Reduce the fraction \frac{-102}{108} to lowest terms by extracting and canceling out 6.

$$x=-\frac{17}{18}$$

The equation is now solved.

$$x=\frac{4}{3} x=-\frac{17}{18}$$

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.

$$54x^{2}-21x-68=0$$

Add 68 to both sides of the equation.

$$54x^{2}-21x-68-\left(-68\right)=-\left(-68\right)$$

Subtracting -68 from itself leaves 0.

$$54x^{2}-21x=-\left(-68\right)$$

Subtract -68 from 0.

$$54x^{2}-21x=68$$

Divide both sides by 54.

$$\frac{54x^{2}-21x}{54}=\frac{68}{54}$$

Dividing by 54 undoes the multiplication by 54.

$$x^{2}+\left(-\frac{21}{54}\right)x=\frac{68}{54}$$

Reduce the fraction \frac{-21}{54} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{7}{18}x=\frac{68}{54}$$

Reduce the fraction \frac{68}{54} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{7}{18}x=\frac{34}{27}$$

Divide -\frac{7}{18}, the coefficient of the x term, by 2 to get -\frac{7}{36}. Then add the square of -\frac{7}{36} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{18}x+\left(-\frac{7}{36}\right)^{2}=\frac{34}{27}+\left(-\frac{7}{36}\right)^{2}$$

Square -\frac{7}{36} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{18}x+\frac{49}{1296}=\frac{34}{27}+\frac{49}{1296}$$

Add \frac{34}{27} to \frac{49}{1296} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{7}{18}x+\frac{49}{1296}=\frac{1681}{1296}$$

Factor x^{2}-\frac{7}{18}x+\frac{49}{1296}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{36}\right)^{2}=\frac{1681}{1296}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{36}\right)^{2}}=\sqrt{\frac{1681}{1296}}$$

Simplify.

$$x-\frac{7}{36}=\frac{41}{36} x-\frac{7}{36}=-\frac{41}{36}$$

Add \frac{7}{36} to both sides of the equation.

$$x=\frac{4}{3} x=-\frac{17}{18}$$