Solve 15y^2=41y-14

$$15 y ^ { 2 } = 41 y - 14$$

Solve for y

$$y=\frac{2}{5}=0.4$$

$$y = \frac{7}{3} = 2\frac{1}{3} \approx 2.333333333$$

Steps Using Factoring By Grouping

Subtract 41y from both sides.

$$15y^{2}-41y=-14$$

Add 14 to both sides.

$$15y^{2}-41y+14=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 15y^{2}+ay+by+14. To find a and b, set up a system to be solved.

$$a+b=-41 ab=15\times 14=210$$

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 210.

$$-1,-210 -2,-105 -3,-70 -5,-42 -6,-35 -7,-30 -10,-21 -14,-15$$

Calculate the sum for each pair.

$$-1-210=-211 -2-105=-107 -3-70=-73 -5-42=-47 -6-35=-41 -7-30=-37 -10-21=-31 -14-15=-29$$

The solution is the pair that gives sum -41.

$$a=-35 b=-6$$

Rewrite 15y^{2}-41y+14 as \left(15y^{2}-35y\right)+\left(-6y+14\right).

$$\left(15y^{2}-35y\right)+\left(-6y+14\right)$$

Factor out 5y in the first and -2 in the second group.

$$5y\left(3y-7\right)-2\left(3y-7\right)$$

Factor out common term 3y-7 by using distributive property.

$$\left(3y-7\right)\left(5y-2\right)$$

To find equation solutions, solve 3y-7=0 and 5y-2=0.

$$y=\frac{7}{3} y=\frac{2}{5}$$

Steps Using the Quadratic Formula

Subtract 41y from both sides.

$$15y^{2}-41y=-14$$

Add 14 to both sides.

$$15y^{2}-41y+14=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 15 for a, -41 for b, and 14 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-41\right)±\sqrt{\left(-41\right)^{2}-4\times 15\times 14}}{2\times 15}$$

Square -41.

$$y=\frac{-\left(-41\right)±\sqrt{1681-4\times 15\times 14}}{2\times 15}$$

Multiply -4 times 15.

$$y=\frac{-\left(-41\right)±\sqrt{1681-60\times 14}}{2\times 15}$$

Multiply -60 times 14.

$$y=\frac{-\left(-41\right)±\sqrt{1681-840}}{2\times 15}$$

Add 1681 to -840.

$$y=\frac{-\left(-41\right)±\sqrt{841}}{2\times 15}$$

Take the square root of 841.

$$y=\frac{-\left(-41\right)±29}{2\times 15}$$

The opposite of -41 is 41.

$$y=\frac{41±29}{2\times 15}$$

Multiply 2 times 15.

$$y=\frac{41±29}{30}$$

Now solve the equation y=\frac{41±29}{30} when ± is plus. Add 41 to 29.

$$y=\frac{70}{30}$$

Reduce the fraction \frac{70}{30} to lowest terms by extracting and canceling out 10.

$$y=\frac{7}{3}$$

Now solve the equation y=\frac{41±29}{30} when ± is minus. Subtract 29 from 41.

$$y=\frac{12}{30}$$

Reduce the fraction \frac{12}{30} to lowest terms by extracting and canceling out 6.

$$y=\frac{2}{5}$$

The equation is now solved.

$$y=\frac{7}{3} y=\frac{2}{5}$$

Steps for Completing the Square

Subtract 41y from both sides.

$$15y^{2}-41y=-14$$

Divide both sides by 15.

$$\frac{15y^{2}-41y}{15}=-\frac{14}{15}$$

Dividing by 15 undoes the multiplication by 15.

$$y^{2}-\frac{41}{15}y=-\frac{14}{15}$$

Divide -\frac{41}{15}, the coefficient of the x term, by 2 to get -\frac{41}{30}. Then add the square of -\frac{41}{30} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-\frac{41}{15}y+\left(-\frac{41}{30}\right)^{2}=-\frac{14}{15}+\left(-\frac{41}{30}\right)^{2}$$

Square -\frac{41}{30} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-\frac{41}{15}y+\frac{1681}{900}=-\frac{14}{15}+\frac{1681}{900}$$

Add -\frac{14}{15} to \frac{1681}{900} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}-\frac{41}{15}y+\frac{1681}{900}=\frac{841}{900}$$

Factor y^{2}-\frac{41}{15}y+\frac{1681}{900}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-\frac{41}{30}\right)^{2}=\frac{841}{900}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{41}{30}\right)^{2}}=\sqrt{\frac{841}{900}}$$

Simplify.

$$y-\frac{41}{30}=\frac{29}{30} y-\frac{41}{30}=-\frac{29}{30}$$

Add \frac{41}{30} to both sides of the equation.

$$y=\frac{7}{3} y=\frac{2}{5}$$