Solve 3y-2=1-6sqrt{y+3}+y+3

$$3 y - 2 = 1 - 6 \sqrt { y + 3 } + y + 3$$

Solve for y

$$y=\frac{15-3\sqrt{33}}{2}\approx -1.11684397$$

Solution Steps

Subtract 1-6\sqrt{y+3} from both sides.

$$3y-2-\left(1-6\sqrt{y+3}\right)=y+3$$

Subtract y from both sides.

$$3y-2-\left(1-6\sqrt{y+3}\right)-y=3$$

To find the opposite of 1-6\sqrt{y+3}, find the opposite of each term.

$$3y-2-1-\left(-6\sqrt{y+3}\right)-y=3$$

The opposite of -6\sqrt{y+3} is 6\sqrt{y+3}.

$$3y-2-1+6\sqrt{y+3}-y=3$$

Subtract 1 from -2 to get -3.

$$3y-3+6\sqrt{y+3}-y=3$$

Combine 3y and -y to get 2y.

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

Add 3 to both sides.

$$2y+6\sqrt{y+3}=3+3$$

Add 3 and 3 to get 6.

$$2y+6\sqrt{y+3}=6$$

Subtract 2y from both sides of the equation.

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

Square both sides of the equation.

$$\left(6\sqrt{y+3}\right)^{2}=\left(6-2y\right)^{2}$$

Expand \left(6\sqrt{y+3}\right)^{2}.

$$6^{2}\left(\sqrt{y+3}\right)^{2}=\left(6-2y\right)^{2}$$

Calculate 6 to the power of 2 and get 36.

$$36\left(\sqrt{y+3}\right)^{2}=\left(6-2y\right)^{2}$$

Calculate \sqrt{y+3} to the power of 2 and get y+3.

$$36\left(y+3\right)=\left(6-2y\right)^{2}$$

Use the distributive property to multiply 36 by y+3.

$$36y+108=\left(6-2y\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6-2y\right)^{2}.

$$36y+108=36-24y+4y^{2}$$

Add 24y to both sides.

$$36y+108+24y=36+4y^{2}$$

Combine 36y and 24y to get 60y.

$$60y+108=36+4y^{2}$$

Subtract 4y^{2} from both sides.

$$60y+108-4y^{2}=36$$

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.

$$-4y^{2}+60y+108=36$$

Subtract 36 from both sides of the equation.

$$-4y^{2}+60y+108-36=36-36$$

Subtracting 36 from itself leaves 0.

$$-4y^{2}+60y+108-36=0$$

Subtract 36 from 108.

$$-4y^{2}+60y+72=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 60 for b, and 72 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-60±\sqrt{60^{2}-4\left(-4\right)\times 72}}{2\left(-4\right)}$$

Square 60.

$$y=\frac{-60±\sqrt{3600-4\left(-4\right)\times 72}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$y=\frac{-60±\sqrt{3600+16\times 72}}{2\left(-4\right)}$$

Multiply 16 times 72.

$$y=\frac{-60±\sqrt{3600+1152}}{2\left(-4\right)}$$

Add 3600 to 1152.

$$y=\frac{-60±\sqrt{4752}}{2\left(-4\right)}$$

Take the square root of 4752.

$$y=\frac{-60±12\sqrt{33}}{2\left(-4\right)}$$

Multiply 2 times -4.

$$y=\frac{-60±12\sqrt{33}}{-8}$$

Now solve the equation y=\frac{-60±12\sqrt{33}}{-8} when ± is plus. Add -60 to 12\sqrt{33}.

$$y=\frac{12\sqrt{33}-60}{-8}$$

Divide -60+12\sqrt{33} by -8.

$$y=\frac{15-3\sqrt{33}}{2}$$

Now solve the equation y=\frac{-60±12\sqrt{33}}{-8} when ± is minus. Subtract 12\sqrt{33} from -60.

$$y=\frac{-12\sqrt{33}-60}{-8}$$

Divide -60-12\sqrt{33} by -8.

$$y=\frac{3\sqrt{33}+15}{2}$$

The equation is now solved.

$$y=\frac{15-3\sqrt{33}}{2} y=\frac{3\sqrt{33}+15}{2}$$

Substitute \frac{15-3\sqrt{33}}{2} for y in the equation 3y-2=1-6\sqrt{y+3}+y+3.

$$3\times \frac{15-3\sqrt{33}}{2}-2=1-6\sqrt{\frac{15-3\sqrt{33}}{2}+3}+\frac{15-3\sqrt{33}}{2}+3$$

Simplify. The value y=\frac{15-3\sqrt{33}}{2} satisfies the equation.

$$\frac{41}{2}-\frac{9}{2}\times 33^{\frac{1}{2}}=\frac{41}{2}-\frac{9}{2}\times 33^{\frac{1}{2}}$$

Substitute \frac{3\sqrt{33}+15}{2} for y in the equation 3y-2=1-6\sqrt{y+3}+y+3.

$$3\times \frac{3\sqrt{33}+15}{2}-2=1-6\sqrt{\frac{3\sqrt{33}+15}{2}+3}+\frac{3\sqrt{33}+15}{2}+3$$

Simplify. The value y=\frac{3\sqrt{33}+15}{2} does not satisfy the equation because the left and the right hand side have opposite signs.

$$\frac{9}{2}\times 33^{\frac{1}{2}}+\frac{41}{2}=\frac{5}{2}-\frac{3}{2}\times 33^{\frac{1}{2}}$$

Equation 6\sqrt{y+3}=6-2y has a unique solution.

$$y=\frac{15-3\sqrt{33}}{2}$$