Solve {left(14-4y/3right)}^2-10(14-4y/3)+y^2-12y-39=0

$${ \left( \frac{ 14-4y }{ 3 } \right) }^{ 2 } -10( \frac{ 14-4y }{ 3 } )+ { y }^{ 2 } -12y-39=0$$

Solve for y

$$y=3\sqrt{3}+2\approx 7.196152423$$

$$y=2-3\sqrt{3}\approx -3.196152423$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 3.

$$3\left(\left(\frac{14-4y}{3}\right)^{2}-10\times \frac{14-4y}{3}\right)+3y^{2}-36y-117=0$$

To raise \frac{14-4y}{3} to a power, raise both numerator and denominator to the power and then divide.

$$3\left(\frac{\left(14-4y\right)^{2}}{3^{2}}-10\times \frac{14-4y}{3}\right)+3y^{2}-36y-117=0$$

Express 10\times \frac{14-4y}{3} as a single fraction.

$$3\left(\frac{\left(14-4y\right)^{2}}{3^{2}}-\frac{10\left(14-4y\right)}{3}\right)+3y^{2}-36y-117=0$$

Use the distributive property to multiply 10 by 14-4y.

$$3\left(\frac{\left(14-4y\right)^{2}}{3^{2}}-\frac{140-40y}{3}\right)+3y^{2}-36y-117=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3^{2} and 3 is 9. Multiply \frac{140-40y}{3} times \frac{3}{3}.

$$3\left(\frac{\left(14-4y\right)^{2}}{9}-\frac{3\left(140-40y\right)}{9}\right)+3y^{2}-36y-117=0$$

Since \frac{\left(14-4y\right)^{2}}{9} and \frac{3\left(140-40y\right)}{9} have the same denominator, subtract them by subtracting their numerators.

$$3\times \frac{\left(14-4y\right)^{2}-3\left(140-40y\right)}{9}+3y^{2}-36y-117=0$$

Do the multiplications in \left(14-4y\right)^{2}-3\left(140-40y\right).

$$3\times \frac{196-112y+16y^{2}-420+120y}{9}+3y^{2}-36y-117=0$$

Combine like terms in 196-112y+16y^{2}-420+120y.

$$3\times \frac{-224+8y+16y^{2}}{9}+3y^{2}-36y-117=0$$

Cancel out 9, the greatest common factor in 3 and 9.

$$\frac{-224+8y+16y^{2}}{3}+3y^{2}-36y-117=0$$

Divide each term of -224+8y+16y^{2} by 3 to get -\frac{224}{3}+\frac{8}{3}y+\frac{16}{3}y^{2}.

$$-\frac{224}{3}+\frac{8}{3}y+\frac{16}{3}y^{2}+3y^{2}-36y-117=0$$

Combine \frac{16}{3}y^{2} and 3y^{2} to get \frac{25}{3}y^{2}.

$$-\frac{224}{3}+\frac{8}{3}y+\frac{25}{3}y^{2}-36y-117=0$$

Combine \frac{8}{3}y and -36y to get -\frac{100}{3}y.

$$-\frac{224}{3}-\frac{100}{3}y+\frac{25}{3}y^{2}-117=0$$

Subtract 117 from -\frac{224}{3} to get -\frac{575}{3}.

$$-\frac{575}{3}-\frac{100}{3}y+\frac{25}{3}y^{2}=0$$

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.

$$\frac{25}{3}y^{2}-\frac{100}{3}y-\frac{575}{3}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{25}{3} for a, -\frac{100}{3} for b, and -\frac{575}{3} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-\frac{100}{3}\right)±\sqrt{\left(-\frac{100}{3}\right)^{2}-4\times \frac{25}{3}\left(-\frac{575}{3}\right)}}{2\times \frac{25}{3}}$$

Square -\frac{100}{3} by squaring both the numerator and the denominator of the fraction.

$$y=\frac{-\left(-\frac{100}{3}\right)±\sqrt{\frac{10000}{9}-4\times \frac{25}{3}\left(-\frac{575}{3}\right)}}{2\times \frac{25}{3}}$$

Multiply -4 times \frac{25}{3}.

$$y=\frac{-\left(-\frac{100}{3}\right)±\sqrt{\frac{10000}{9}-\frac{100}{3}\left(-\frac{575}{3}\right)}}{2\times \frac{25}{3}}$$

Multiply -\frac{100}{3} times -\frac{575}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\left(-\frac{100}{3}\right)±\sqrt{\frac{10000+57500}{9}}}{2\times \frac{25}{3}}$$

Add \frac{10000}{9} to \frac{57500}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\left(-\frac{100}{3}\right)±\sqrt{7500}}{2\times \frac{25}{3}}$$

Take the square root of 7500.

$$y=\frac{-\left(-\frac{100}{3}\right)±50\sqrt{3}}{2\times \frac{25}{3}}$$

The opposite of -\frac{100}{3} is \frac{100}{3}.

$$y=\frac{\frac{100}{3}±50\sqrt{3}}{2\times \frac{25}{3}}$$

Multiply 2 times \frac{25}{3}.

$$y=\frac{\frac{100}{3}±50\sqrt{3}}{\frac{50}{3}}$$

Now solve the equation y=\frac{\frac{100}{3}±50\sqrt{3}}{\frac{50}{3}} when ± is plus. Add \frac{100}{3} to 50\sqrt{3}.

$$y=\frac{50\sqrt{3}+\frac{100}{3}}{\frac{50}{3}}$$

Divide \frac{100}{3}+50\sqrt{3} by \frac{50}{3} by multiplying \frac{100}{3}+50\sqrt{3} by the reciprocal of \frac{50}{3}.

$$y=3\sqrt{3}+2$$

Now solve the equation y=\frac{\frac{100}{3}±50\sqrt{3}}{\frac{50}{3}} when ± is minus. Subtract 50\sqrt{3} from \frac{100}{3}.

$$y=\frac{\frac{100}{3}-50\sqrt{3}}{\frac{50}{3}}$$

Divide \frac{100}{3}-50\sqrt{3} by \frac{50}{3} by multiplying \frac{100}{3}-50\sqrt{3} by the reciprocal of \frac{50}{3}.

$$y=2-3\sqrt{3}$$

The equation is now solved.

$$y=3\sqrt{3}+2 y=2-3\sqrt{3}$$

Steps for Completing the Square

Multiply both sides of the equation by 3.

$$3\left(\left(\frac{14-4y}{3}\right)^{2}-10\times \frac{14-4y}{3}\right)+3y^{2}-36y-117=0$$

To raise \frac{14-4y}{3} to a power, raise both numerator and denominator to the power and then divide.

$$3\left(\frac{\left(14-4y\right)^{2}}{3^{2}}-10\times \frac{14-4y}{3}\right)+3y^{2}-36y-117=0$$

Express 10\times \frac{14-4y}{3} as a single fraction.

$$3\left(\frac{\left(14-4y\right)^{2}}{3^{2}}-\frac{10\left(14-4y\right)}{3}\right)+3y^{2}-36y-117=0$$

Use the distributive property to multiply 10 by 14-4y.

$$3\left(\frac{\left(14-4y\right)^{2}}{3^{2}}-\frac{140-40y}{3}\right)+3y^{2}-36y-117=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3^{2} and 3 is 9. Multiply \frac{140-40y}{3} times \frac{3}{3}.

$$3\left(\frac{\left(14-4y\right)^{2}}{9}-\frac{3\left(140-40y\right)}{9}\right)+3y^{2}-36y-117=0$$

Since \frac{\left(14-4y\right)^{2}}{9} and \frac{3\left(140-40y\right)}{9} have the same denominator, subtract them by subtracting their numerators.

$$3\times \frac{\left(14-4y\right)^{2}-3\left(140-40y\right)}{9}+3y^{2}-36y-117=0$$

Do the multiplications in \left(14-4y\right)^{2}-3\left(140-40y\right).

$$3\times \frac{196-112y+16y^{2}-420+120y}{9}+3y^{2}-36y-117=0$$

Combine like terms in 196-112y+16y^{2}-420+120y.

$$3\times \frac{-224+8y+16y^{2}}{9}+3y^{2}-36y-117=0$$

Cancel out 9, the greatest common factor in 3 and 9.

$$\frac{-224+8y+16y^{2}}{3}+3y^{2}-36y-117=0$$

Divide each term of -224+8y+16y^{2} by 3 to get -\frac{224}{3}+\frac{8}{3}y+\frac{16}{3}y^{2}.

$$-\frac{224}{3}+\frac{8}{3}y+\frac{16}{3}y^{2}+3y^{2}-36y-117=0$$

Combine \frac{16}{3}y^{2} and 3y^{2} to get \frac{25}{3}y^{2}.

$$-\frac{224}{3}+\frac{8}{3}y+\frac{25}{3}y^{2}-36y-117=0$$

Combine \frac{8}{3}y and -36y to get -\frac{100}{3}y.

$$-\frac{224}{3}-\frac{100}{3}y+\frac{25}{3}y^{2}-117=0$$

Subtract 117 from -\frac{224}{3} to get -\frac{575}{3}.

$$-\frac{575}{3}-\frac{100}{3}y+\frac{25}{3}y^{2}=0$$

Add \frac{575}{3} to both sides. Anything plus zero gives itself.

$$-\frac{100}{3}y+\frac{25}{3}y^{2}=\frac{575}{3}$$

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.

$$\frac{25}{3}y^{2}-\frac{100}{3}y=\frac{575}{3}$$

Divide both sides of the equation by \frac{25}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{25}{3}y^{2}-\frac{100}{3}y}{\frac{25}{3}}=\frac{\frac{575}{3}}{\frac{25}{3}}$$

Dividing by \frac{25}{3} undoes the multiplication by \frac{25}{3}.

$$y^{2}+\left(-\frac{\frac{100}{3}}{\frac{25}{3}}\right)y=\frac{\frac{575}{3}}{\frac{25}{3}}$$

Divide -\frac{100}{3} by \frac{25}{3} by multiplying -\frac{100}{3} by the reciprocal of \frac{25}{3}.

$$y^{2}-4y=\frac{\frac{575}{3}}{\frac{25}{3}}$$

Divide \frac{575}{3} by \frac{25}{3} by multiplying \frac{575}{3} by the reciprocal of \frac{25}{3}.

$$y^{2}-4y=23$$

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-4y+\left(-2\right)^{2}=23+\left(-2\right)^{2}$$

Square -2.

$$y^{2}-4y+4=23+4$$

Add 23 to 4.

$$y^{2}-4y+4=27$$

Factor y^{2}-4y+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-2\right)^{2}=27$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-2\right)^{2}}=\sqrt{27}$$

Simplify.

$$y-2=3\sqrt{3} y-2=-3\sqrt{3}$$

Add 2 to both sides of the equation.

$$y=3\sqrt{3}+2 y=2-3\sqrt{3}$$