Solve 3x^2-41x=-110

$$3 x ^ { 2 } - 41 x = - 110$$

Solve for x

$$x=10$$

$$x = \frac{11}{3} = 3\frac{2}{3} \approx 3.666666667$$

Steps Using Factoring By Grouping

Add 110 to both sides.

$$3x^{2}-41x+110=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx+110. To find a and b, set up a system to be solved.

$$a+b=-41 ab=3\times 110=330$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 330.

$$-1,-330 -2,-165 -3,-110 -5,-66 -6,-55 -10,-33 -11,-30 -15,-22$$

Calculate the sum for each pair.

$$-1-330=-331 -2-165=-167 -3-110=-113 -5-66=-71 -6-55=-61 -10-33=-43 -11-30=-41 -15-22=-37$$

The solution is the pair that gives sum -41.

$$a=-30 b=-11$$

Rewrite 3x^{2}-41x+110 as \left(3x^{2}-30x\right)+\left(-11x+110\right).

$$\left(3x^{2}-30x\right)+\left(-11x+110\right)$$

Factor out 3x in the first and -11 in the second group.

$$3x\left(x-10\right)-11\left(x-10\right)$$

Factor out common term x-10 by using distributive property.

$$\left(x-10\right)\left(3x-11\right)$$

To find equation solutions, solve x-10=0 and 3x-11=0.

$$x=10 x=\frac{11}{3}$$

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.

$$3x^{2}-41x=-110$$

Add 110 to both sides of the equation.

$$3x^{2}-41x-\left(-110\right)=-110-\left(-110\right)$$

Subtracting -110 from itself leaves 0.

$$3x^{2}-41x-\left(-110\right)=0$$

Subtract -110 from 0.

$$3x^{2}-41x+110=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -41 for b, and 110 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-41\right)±\sqrt{\left(-41\right)^{2}-4\times 3\times 110}}{2\times 3}$$

Square -41.

$$x=\frac{-\left(-41\right)±\sqrt{1681-4\times 3\times 110}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-41\right)±\sqrt{1681-12\times 110}}{2\times 3}$$

Multiply -12 times 110.

$$x=\frac{-\left(-41\right)±\sqrt{1681-1320}}{2\times 3}$$

Add 1681 to -1320.

$$x=\frac{-\left(-41\right)±\sqrt{361}}{2\times 3}$$

Take the square root of 361.

$$x=\frac{-\left(-41\right)±19}{2\times 3}$$

The opposite of -41 is 41.

$$x=\frac{41±19}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{41±19}{6}$$

Now solve the equation x=\frac{41±19}{6} when ± is plus. Add 41 to 19.

$$x=\frac{60}{6}$$

Divide 60 by 6.

$$x=10$$

Now solve the equation x=\frac{41±19}{6} when ± is minus. Subtract 19 from 41.

$$x=\frac{22}{6}$$

Reduce the fraction \frac{22}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{11}{3}$$

The equation is now solved.

$$x=10 x=\frac{11}{3}$$

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.

$$3x^{2}-41x=-110$$

Divide both sides by 3.

$$\frac{3x^{2}-41x}{3}=-\frac{110}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{41}{3}x=-\frac{110}{3}$$

Divide -\frac{41}{3}, the coefficient of the x term, by 2 to get -\frac{41}{6}. Then add the square of -\frac{41}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{41}{3}x+\left(-\frac{41}{6}\right)^{2}=-\frac{110}{3}+\left(-\frac{41}{6}\right)^{2}$$

Square -\frac{41}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{41}{3}x+\frac{1681}{36}=-\frac{110}{3}+\frac{1681}{36}$$

Add -\frac{110}{3} to \frac{1681}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{41}{3}x+\frac{1681}{36}=\frac{361}{36}$$

Factor x^{2}-\frac{41}{3}x+\frac{1681}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{41}{6}\right)^{2}=\frac{361}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{41}{6}\right)^{2}}=\sqrt{\frac{361}{36}}$$

Simplify.

$$x-\frac{41}{6}=\frac{19}{6} x-\frac{41}{6}=-\frac{19}{6}$$

Add \frac{41}{6} to both sides of the equation.

$$x=10 x=\frac{11}{3}$$