Solve (y+5)^2=2y^2+5y+19

$$( y + 5 ) ^ { 2 } = 2 y ^ { 2 } + 5 y + 19$$

Solve for y

$$y=-1$$

$$y=6$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+5\right)^{2}.

$$y^{2}+10y+25=2y^{2}+5y+19$$

Subtract 2y^{2} from both sides.

$$y^{2}+10y+25-2y^{2}=5y+19$$

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

$$-y^{2}+10y+25=5y+19$$

Subtract 5y from both sides.

$$-y^{2}+10y+25-5y=19$$

Combine 10y and -5y to get 5y.

$$-y^{2}+5y+25=19$$

Subtract 19 from both sides.

$$-y^{2}+5y+25-19=0$$

Subtract 19 from 25 to get 6.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -y^{2}+ay+by+6. To find a and b, set up a system to be solved.

$$a+b=5 ab=-6=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -6.

$$-1,6 -2,3$$

Calculate the sum for each pair.

$$-1+6=5 -2+3=1$$

The solution is the pair that gives sum 5.

$$a=6 b=-1$$

Rewrite -y^{2}+5y+6 as \left(-y^{2}+6y\right)+\left(-y+6\right).

$$\left(-y^{2}+6y\right)+\left(-y+6\right)$$

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

$$-y\left(y-6\right)-\left(y-6\right)$$

Factor out common term y-6 by using distributive property.

$$\left(y-6\right)\left(-y-1\right)$$

To find equation solutions, solve y-6=0 and -y-1=0.

$$y=6 y=-1$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+5\right)^{2}.

$$y^{2}+10y+25=2y^{2}+5y+19$$

Subtract 2y^{2} from both sides.

$$y^{2}+10y+25-2y^{2}=5y+19$$

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

$$-y^{2}+10y+25=5y+19$$

Subtract 5y from both sides.

$$-y^{2}+10y+25-5y=19$$

Combine 10y and -5y to get 5y.

$$-y^{2}+5y+25=19$$

Subtract 19 from both sides.

$$-y^{2}+5y+25-19=0$$

Subtract 19 from 25 to get 6.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 5 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-5±\sqrt{5^{2}-4\left(-1\right)\times 6}}{2\left(-1\right)}$$

Square 5.

$$y=\frac{-5±\sqrt{25-4\left(-1\right)\times 6}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$y=\frac{-5±\sqrt{25+4\times 6}}{2\left(-1\right)}$$

Multiply 4 times 6.

$$y=\frac{-5±\sqrt{25+24}}{2\left(-1\right)}$$

Add 25 to 24.

$$y=\frac{-5±\sqrt{49}}{2\left(-1\right)}$$

Take the square root of 49.

$$y=\frac{-5±7}{2\left(-1\right)}$$

Multiply 2 times -1.

$$y=\frac{-5±7}{-2}$$

Now solve the equation y=\frac{-5±7}{-2} when ± is plus. Add -5 to 7.

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

Divide 2 by -2.

$$y=-1$$

Now solve the equation y=\frac{-5±7}{-2} when ± is minus. Subtract 7 from -5.

$$y=-\frac{12}{-2}$$

Divide -12 by -2.

$$y=6$$

The equation is now solved.

$$y=-1 y=6$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+5\right)^{2}.

$$y^{2}+10y+25=2y^{2}+5y+19$$

Subtract 2y^{2} from both sides.

$$y^{2}+10y+25-2y^{2}=5y+19$$

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

$$-y^{2}+10y+25=5y+19$$

Subtract 5y from both sides.

$$-y^{2}+10y+25-5y=19$$

Combine 10y and -5y to get 5y.

$$-y^{2}+5y+25=19$$

Subtract 25 from both sides.

$$-y^{2}+5y=19-25$$

Subtract 25 from 19 to get -6.

$$-y^{2}+5y=-6$$

Divide both sides by -1.

$$\frac{-y^{2}+5y}{-1}=-\frac{6}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$y^{2}+\frac{5}{-1}y=-\frac{6}{-1}$$

Divide 5 by -1.

$$y^{2}-5y=-\frac{6}{-1}$$

Divide -6 by -1.

$$y^{2}-5y=6$$

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-5y+\left(-\frac{5}{2}\right)^{2}=6+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-5y+\frac{25}{4}=6+\frac{25}{4}$$

Add 6 to \frac{25}{4}.

$$y^{2}-5y+\frac{25}{4}=\frac{49}{4}$$

Factor y^{2}-5y+\frac{25}{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}.

$$\left(y-\frac{5}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{5}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$y-\frac{5}{2}=\frac{7}{2} y-\frac{5}{2}=-\frac{7}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$y=6 y=-1$$