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

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

Solve for y

$$y = \frac{\sqrt{37} + 3}{2} \approx 4.541381265$$

$$y=\frac{3-\sqrt{37}}{2}\approx -1.541381265$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-4\right)^{2}.

$$y^{2}-8y+16+5\left(y-4\right)-3=0$$

Use the distributive property to multiply 5 by y-4.

$$y^{2}-8y+16+5y-20-3=0$$

Combine -8y and 5y to get -3y.

$$y^{2}-3y+16-20-3=0$$

Subtract 20 from 16 to get -4.

$$y^{2}-3y-4-3=0$$

Subtract 3 from -4 to get -7.

$$y^{2}-3y-7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -3 for b, and -7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-7\right)}}{2}$$

Square -3.

$$y=\frac{-\left(-3\right)±\sqrt{9-4\left(-7\right)}}{2}$$

Multiply -4 times -7.

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

Add 9 to 28.

$$y=\frac{-\left(-3\right)±\sqrt{37}}{2}$$

The opposite of -3 is 3.

$$y=\frac{3±\sqrt{37}}{2}$$

Now solve the equation y=\frac{3±\sqrt{37}}{2} when ± is plus. Add 3 to \sqrt{37}.

$$y=\frac{\sqrt{37}+3}{2}$$

Now solve the equation y=\frac{3±\sqrt{37}}{2} when ± is minus. Subtract \sqrt{37} from 3.

$$y=\frac{3-\sqrt{37}}{2}$$

The equation is now solved.

$$y=\frac{\sqrt{37}+3}{2} y=\frac{3-\sqrt{37}}{2}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-4\right)^{2}.

$$y^{2}-8y+16+5\left(y-4\right)-3=0$$

Use the distributive property to multiply 5 by y-4.

$$y^{2}-8y+16+5y-20-3=0$$

Combine -8y and 5y to get -3y.

$$y^{2}-3y+16-20-3=0$$

Subtract 20 from 16 to get -4.

$$y^{2}-3y-4-3=0$$

Subtract 3 from -4 to get -7.

$$y^{2}-3y-7=0$$

Add 7 to both sides. Anything plus zero gives itself.

$$y^{2}-3y=7$$

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.

$$y^{2}-3y+\left(-\frac{3}{2}\right)^{2}=7+\left(-\frac{3}{2}\right)^{2}$$

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

$$y^{2}-3y+\frac{9}{4}=7+\frac{9}{4}$$

Add 7 to \frac{9}{4}.

$$y^{2}-3y+\frac{9}{4}=\frac{37}{4}$$

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}.

$$\left(y-\frac{3}{2}\right)^{2}=\frac{37}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{3}{2}\right)^{2}}=\sqrt{\frac{37}{4}}$$

Simplify.

$$y-\frac{3}{2}=\frac{\sqrt{37}}{2} y-\frac{3}{2}=-\frac{\sqrt{37}}{2}$$

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

$$y=\frac{\sqrt{37}+3}{2} y=\frac{3-\sqrt{37}}{2}$$