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

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

Solve for y

$$y=1$$

$$y = \frac{9}{2} = 4\frac{1}{2} = 4.5$$

Steps Using Factoring By Grouping

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

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

Use the distributive property to multiply 2 by y^{2}-8y+16.

$$2y^{2}-16y+32+5\left(y-4\right)-3=0$$

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

$$2y^{2}-16y+32+5y-20-3=0$$

Combine -16y and 5y to get -11y.

$$2y^{2}-11y+32-20-3=0$$

Subtract 20 from 32 to get 12.

$$2y^{2}-11y+12-3=0$$

Subtract 3 from 12 to get 9.

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

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

$$a+b=-11 ab=2\times 9=18$$

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 18.

$$-1,-18 -2,-9 -3,-6$$

Calculate the sum for each pair.

$$-1-18=-19 -2-9=-11 -3-6=-9$$

The solution is the pair that gives sum -11.

$$a=-9 b=-2$$

Rewrite 2y^{2}-11y+9 as \left(2y^{2}-9y\right)+\left(-2y+9\right).

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

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

$$y\left(2y-9\right)-\left(2y-9\right)$$

Factor out common term 2y-9 by using distributive property.

$$\left(2y-9\right)\left(y-1\right)$$

To find equation solutions, solve 2y-9=0 and y-1=0.

$$y=\frac{9}{2} y=1$$

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}.

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

Use the distributive property to multiply 2 by y^{2}-8y+16.

$$2y^{2}-16y+32+5\left(y-4\right)-3=0$$

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

$$2y^{2}-16y+32+5y-20-3=0$$

Combine -16y and 5y to get -11y.

$$2y^{2}-11y+32-20-3=0$$

Subtract 20 from 32 to get 12.

$$2y^{2}-11y+12-3=0$$

Subtract 3 from 12 to get 9.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -11 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 2\times 9}}{2\times 2}$$

Square -11.

$$y=\frac{-\left(-11\right)±\sqrt{121-4\times 2\times 9}}{2\times 2}$$

Multiply -4 times 2.

$$y=\frac{-\left(-11\right)±\sqrt{121-8\times 9}}{2\times 2}$$

Multiply -8 times 9.

$$y=\frac{-\left(-11\right)±\sqrt{121-72}}{2\times 2}$$

Add 121 to -72.

$$y=\frac{-\left(-11\right)±\sqrt{49}}{2\times 2}$$

Take the square root of 49.

$$y=\frac{-\left(-11\right)±7}{2\times 2}$$

The opposite of -11 is 11.

$$y=\frac{11±7}{2\times 2}$$

Multiply 2 times 2.

$$y=\frac{11±7}{4}$$

Now solve the equation y=\frac{11±7}{4} when ± is plus. Add 11 to 7.

$$y=\frac{18}{4}$$

Reduce the fraction \frac{18}{4} to lowest terms by extracting and canceling out 2.

$$y=\frac{9}{2}$$

Now solve the equation y=\frac{11±7}{4} when ± is minus. Subtract 7 from 11.

$$y=\frac{4}{4}$$

Divide 4 by 4.

$$y=1$$

The equation is now solved.

$$y=\frac{9}{2} y=1$$

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}.

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

Use the distributive property to multiply 2 by y^{2}-8y+16.

$$2y^{2}-16y+32+5\left(y-4\right)-3=0$$

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

$$2y^{2}-16y+32+5y-20-3=0$$

Combine -16y and 5y to get -11y.

$$2y^{2}-11y+32-20-3=0$$

Subtract 20 from 32 to get 12.

$$2y^{2}-11y+12-3=0$$

Subtract 3 from 12 to get 9.

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

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

$$2y^{2}-11y=-9$$

Divide both sides by 2.

$$\frac{2y^{2}-11y}{2}=-\frac{9}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$y^{2}-\frac{11}{2}y=-\frac{9}{2}$$

Divide -\frac{11}{2}, the coefficient of the x term, by 2 to get -\frac{11}{4}. Then add the square of -\frac{11}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-\frac{11}{2}y+\left(-\frac{11}{4}\right)^{2}=-\frac{9}{2}+\left(-\frac{11}{4}\right)^{2}$$

Square -\frac{11}{4} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-\frac{11}{2}y+\frac{121}{16}=-\frac{9}{2}+\frac{121}{16}$$

Add -\frac{9}{2} to \frac{121}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}-\frac{11}{2}y+\frac{121}{16}=\frac{49}{16}$$

Factor y^{2}-\frac{11}{2}y+\frac{121}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-\frac{11}{4}\right)^{2}=\frac{49}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{11}{4}\right)^{2}}=\sqrt{\frac{49}{16}}$$

Simplify.

$$y-\frac{11}{4}=\frac{7}{4} y-\frac{11}{4}=-\frac{7}{4}$$

Add \frac{11}{4} to both sides of the equation.

$$y=\frac{9}{2} y=1$$