Solve left(y-3right)^2-5(y+3)+4=0

$${ \left(y-3 \right) }^{ 2 } -5(y+3)+4=0$$

Solve for y

$$y = \frac{\sqrt{129} + 11}{2} \approx 11.178908346$$

$$y=\frac{11-\sqrt{129}}{2}\approx -0.178908346$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-3\right)^{2}.

$$y^{2}-6y+9-5\left(y+3\right)+4=0$$

Use the distributive property to multiply -5 by y+3.

$$y^{2}-6y+9-5y-15+4=0$$

Combine -6y and -5y to get -11y.

$$y^{2}-11y+9-15+4=0$$

Subtract 15 from 9 to get -6.

$$y^{2}-11y-6+4=0$$

Add -6 and 4 to get -2.

$$y^{2}-11y-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -11 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\left(-2\right)}}{2}$$

Square -11.

$$y=\frac{-\left(-11\right)±\sqrt{121-4\left(-2\right)}}{2}$$

Multiply -4 times -2.

$$y=\frac{-\left(-11\right)±\sqrt{121+8}}{2}$$

Add 121 to 8.

$$y=\frac{-\left(-11\right)±\sqrt{129}}{2}$$

The opposite of -11 is 11.

$$y=\frac{11±\sqrt{129}}{2}$$

Now solve the equation y=\frac{11±\sqrt{129}}{2} when ± is plus. Add 11 to \sqrt{129}.

$$y=\frac{\sqrt{129}+11}{2}$$

Now solve the equation y=\frac{11±\sqrt{129}}{2} when ± is minus. Subtract \sqrt{129} from 11.

$$y=\frac{11-\sqrt{129}}{2}$$

The equation is now solved.

$$y=\frac{\sqrt{129}+11}{2} y=\frac{11-\sqrt{129}}{2}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-3\right)^{2}.

$$y^{2}-6y+9-5\left(y+3\right)+4=0$$

Use the distributive property to multiply -5 by y+3.

$$y^{2}-6y+9-5y-15+4=0$$

Combine -6y and -5y to get -11y.

$$y^{2}-11y+9-15+4=0$$

Subtract 15 from 9 to get -6.

$$y^{2}-11y-6+4=0$$

Add -6 and 4 to get -2.

$$y^{2}-11y-2=0$$

Add 2 to both sides. Anything plus zero gives itself.

$$y^{2}-11y=2$$

Divide -11, the coefficient of the x term, by 2 to get -\frac{11}{2}. Then add the square of -\frac{11}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-11y+\left(-\frac{11}{2}\right)^{2}=2+\left(-\frac{11}{2}\right)^{2}$$

Square -\frac{11}{2} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-11y+\frac{121}{4}=2+\frac{121}{4}$$

Add 2 to \frac{121}{4}.

$$y^{2}-11y+\frac{121}{4}=\frac{129}{4}$$

Factor y^{2}-11y+\frac{121}{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}.

$$\left(y-\frac{11}{2}\right)^{2}=\frac{129}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{11}{2}\right)^{2}}=\sqrt{\frac{129}{4}}$$

Simplify.

$$y-\frac{11}{2}=\frac{\sqrt{129}}{2} y-\frac{11}{2}=-\frac{\sqrt{129}}{2}$$

Add \frac{11}{2} to both sides of the equation.

$$y=\frac{\sqrt{129}+11}{2} y=\frac{11-\sqrt{129}}{2}$$