Solve 9-6y+3y^2-(3y-y^2)-y^2-1=0

$$9 - 6 y + 3 y ^ { 2 } - ( 3 y - y ^ { 2 } ) - y ^ { 2 } - 1 = 0$$

Solve for y

$$y=\frac{\sqrt{15}i}{6}+\frac{3}{2}\approx 1.5+0.645497224i$$

$$y=-\frac{\sqrt{15}i}{6}+\frac{3}{2}\approx 1.5-0.645497224i$$

Steps Using the Quadratic Formula

To find the opposite of 3y-y^{2}, find the opposite of each term.

$$9-6y+3y^{2}-3y+y^{2}-y^{2}-1=0$$

Combine -6y and -3y to get -9y.

$$9-9y+3y^{2}+y^{2}-y^{2}-1=0$$

Combine 3y^{2} and y^{2} to get 4y^{2}.

$$9-9y+4y^{2}-y^{2}-1=0$$

Combine 4y^{2} and -y^{2} to get 3y^{2}.

$$9-9y+3y^{2}-1=0$$

Subtract 1 from 9 to get 8.

$$8-9y+3y^{2}=0$$

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.

$$3y^{2}-9y+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -9 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 3\times 8}}{2\times 3}$$

Square -9.

$$y=\frac{-\left(-9\right)±\sqrt{81-4\times 3\times 8}}{2\times 3}$$

Multiply -4 times 3.

$$y=\frac{-\left(-9\right)±\sqrt{81-12\times 8}}{2\times 3}$$

Multiply -12 times 8.

$$y=\frac{-\left(-9\right)±\sqrt{81-96}}{2\times 3}$$

Add 81 to -96.

$$y=\frac{-\left(-9\right)±\sqrt{-15}}{2\times 3}$$

Take the square root of -15.

$$y=\frac{-\left(-9\right)±\sqrt{15}i}{2\times 3}$$

The opposite of -9 is 9.

$$y=\frac{9±\sqrt{15}i}{2\times 3}$$

Multiply 2 times 3.

$$y=\frac{9±\sqrt{15}i}{6}$$

Now solve the equation y=\frac{9±\sqrt{15}i}{6} when ± is plus. Add 9 to i\sqrt{15}.

$$y=\frac{9+\sqrt{15}i}{6}$$

Divide 9+i\sqrt{15} by 6.

$$y=\frac{\sqrt{15}i}{6}+\frac{3}{2}$$

Now solve the equation y=\frac{9±\sqrt{15}i}{6} when ± is minus. Subtract i\sqrt{15} from 9.

$$y=\frac{-\sqrt{15}i+9}{6}$$

Divide 9-i\sqrt{15} by 6.

$$y=-\frac{\sqrt{15}i}{6}+\frac{3}{2}$$

The equation is now solved.

$$y=\frac{\sqrt{15}i}{6}+\frac{3}{2} y=-\frac{\sqrt{15}i}{6}+\frac{3}{2}$$

Steps for Completing the Square

To find the opposite of 3y-y^{2}, find the opposite of each term.

$$9-6y+3y^{2}-3y+y^{2}-y^{2}-1=0$$

Combine -6y and -3y to get -9y.

$$9-9y+3y^{2}+y^{2}-y^{2}-1=0$$

Combine 3y^{2} and y^{2} to get 4y^{2}.

$$9-9y+4y^{2}-y^{2}-1=0$$

Combine 4y^{2} and -y^{2} to get 3y^{2}.

$$9-9y+3y^{2}-1=0$$

Subtract 1 from 9 to get 8.

$$8-9y+3y^{2}=0$$

Subtract 8 from both sides. Anything subtracted from zero gives its negation.

$$-9y+3y^{2}=-8$$

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.

$$3y^{2}-9y=-8$$

Divide both sides by 3.

$$\frac{3y^{2}-9y}{3}=-\frac{8}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$y^{2}+\left(-\frac{9}{3}\right)y=-\frac{8}{3}$$

Divide -9 by 3.

$$y^{2}-3y=-\frac{8}{3}$$

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

$$y^{2}-3y+\left(-\frac{3}{2}\right)^{2}=-\frac{8}{3}+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-3y+\frac{9}{4}=-\frac{8}{3}+\frac{9}{4}$$

Add -\frac{8}{3} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}-3y+\frac{9}{4}=-\frac{5}{12}$$

Factor y^{2}-3y+\frac{9}{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}.

$$\left(y-\frac{3}{2}\right)^{2}=-\frac{5}{12}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{3}{2}\right)^{2}}=\sqrt{-\frac{5}{12}}$$

Simplify.

$$y-\frac{3}{2}=\frac{\sqrt{15}i}{6} y-\frac{3}{2}=-\frac{\sqrt{15}i}{6}$$

Add \frac{3}{2} to both sides of the equation.

$$y=\frac{\sqrt{15}i}{6}+\frac{3}{2} y=-\frac{\sqrt{15}i}{6}+\frac{3}{2}$$