Solve 3(y^2+3)=28y

$$3 ( y ^ { 2 } + 3 ) = 28 y$$

Solve for y

$$y=\frac{1}{3}\approx 0.333333333$$

$$y=9$$

Steps Using Factoring By Grouping

Use the distributive property to multiply 3 by y^{2}+3.

$$3y^{2}+9=28y$$

Subtract 28y from both sides.

$$3y^{2}+9-28y=0$$

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

$$3y^{2}-28y+9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3y^{2}+ay+by+9. To find a and b, set up a system to be solved.

$$a+b=-28 ab=3\times 9=27$$

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 27.

$$-1,-27 -3,-9$$

Calculate the sum for each pair.

$$-1-27=-28 -3-9=-12$$

The solution is the pair that gives sum -28.

$$a=-27 b=-1$$

Rewrite 3y^{2}-28y+9 as \left(3y^{2}-27y\right)+\left(-y+9\right).

$$\left(3y^{2}-27y\right)+\left(-y+9\right)$$

Factor out 3y in the first and -1 in the second group.

$$3y\left(y-9\right)-\left(y-9\right)$$

Factor out common term y-9 by using distributive property.

$$\left(y-9\right)\left(3y-1\right)$$

To find equation solutions, solve y-9=0 and 3y-1=0.

$$y=9 y=\frac{1}{3}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3 by y^{2}+3.

$$3y^{2}+9=28y$$

Subtract 28y from both sides.

$$3y^{2}+9-28y=0$$

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.

$$3y^{2}-28y+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -28 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}-4\times 3\times 9}}{2\times 3}$$

Square -28.

$$y=\frac{-\left(-28\right)±\sqrt{784-4\times 3\times 9}}{2\times 3}$$

Multiply -4 times 3.

$$y=\frac{-\left(-28\right)±\sqrt{784-12\times 9}}{2\times 3}$$

Multiply -12 times 9.

$$y=\frac{-\left(-28\right)±\sqrt{784-108}}{2\times 3}$$

Add 784 to -108.

$$y=\frac{-\left(-28\right)±\sqrt{676}}{2\times 3}$$

Take the square root of 676.

$$y=\frac{-\left(-28\right)±26}{2\times 3}$$

The opposite of -28 is 28.

$$y=\frac{28±26}{2\times 3}$$

Multiply 2 times 3.

$$y=\frac{28±26}{6}$$

Now solve the equation y=\frac{28±26}{6} when ± is plus. Add 28 to 26.

$$y=\frac{54}{6}$$

Divide 54 by 6.

$$y=9$$

Now solve the equation y=\frac{28±26}{6} when ± is minus. Subtract 26 from 28.

$$y=\frac{2}{6}$$

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

$$y=\frac{1}{3}$$

The equation is now solved.

$$y=9 y=\frac{1}{3}$$

Steps for Completing the Square

Use the distributive property to multiply 3 by y^{2}+3.

$$3y^{2}+9=28y$$

Subtract 28y from both sides.

$$3y^{2}+9-28y=0$$

Subtract 9 from both sides. Anything subtracted from zero gives its negation.

$$3y^{2}-28y=-9$$

Divide both sides by 3.

$$\frac{3y^{2}-28y}{3}=-\frac{9}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$y^{2}-\frac{28}{3}y=-\frac{9}{3}$$

Divide -9 by 3.

$$y^{2}-\frac{28}{3}y=-3$$

Divide -\frac{28}{3}, the coefficient of the x term, by 2 to get -\frac{14}{3}. Then add the square of -\frac{14}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-\frac{28}{3}y+\left(-\frac{14}{3}\right)^{2}=-3+\left(-\frac{14}{3}\right)^{2}$$

Square -\frac{14}{3} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-\frac{28}{3}y+\frac{196}{9}=-3+\frac{196}{9}$$

Add -3 to \frac{196}{9}.

$$y^{2}-\frac{28}{3}y+\frac{196}{9}=\frac{169}{9}$$

Factor y^{2}-\frac{28}{3}y+\frac{196}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-\frac{14}{3}\right)^{2}=\frac{169}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{14}{3}\right)^{2}}=\sqrt{\frac{169}{9}}$$

Simplify.

$$y-\frac{14}{3}=\frac{13}{3} y-\frac{14}{3}=-\frac{13}{3}$$

Add \frac{14}{3} to both sides of the equation.

$$y=9 y=\frac{1}{3}$$