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\sqrt{166+39y}=14+y
Subtract -y from both sides of the equation.
\left(\sqrt{166+39y}\right)^{2}=\left(14+y\right)^{2}
Square both sides of the equation.
166+39y=\left(14+y\right)^{2}
Calculate \sqrt{166+39y} to the power of 2 and get 166+39y.
166+39y=196+28y+y^{2}
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(14+y\right)^{2}.
166+39y-196=28y+y^{2}
Subtract 196 from both sides.
-30+39y=28y+y^{2}
Subtract 196 from 166 to get -30.
-30+39y-28y=y^{2}
Subtract 28y from both sides.
-30+11y=y^{2}
Combine 39y and -28y to get 11y.
-30+11y-y^{2}=0
Subtract y^{2} from both sides.
-y^{2}+11y-30=0
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
a+b=11 ab=-\left(-30\right)=30
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -y^{2}+ay+by-30. To find a and b, set up a system to be solved.
1,30 2,15 3,10 5,6
Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 30.
1+30=31 2+15=17 3+10=13 5+6=11
Calculate the sum for each pair.
a=6 b=5
The solution is the pair that gives sum 11.
\left(-y^{2}+6y\right)+\left(5y-30\right)
Rewrite -y^{2}+11y-30 as \left(-y^{2}+6y\right)+\left(5y-30\right).
-y\left(y-6\right)+5\left(y-6\right)
Factor out -y in the first and 5 in the second group.
\left(y-6\right)\left(-y+5\right)
Factor out common term y-6 by using distributive property.
y=6 y=5
To find equation solutions, solve y-6=0 and -y+5=0.
\sqrt{166+39\times 6}-6=14
Substitute 6 for y in the equation \sqrt{166+39y}-y=14.
14=14
Simplify. The value y=6 satisfies the equation.
\sqrt{166+39\times 5}-5=14
Substitute 5 for y in the equation \sqrt{166+39y}-y=14.
14=14
Simplify. The value y=5 satisfies the equation.
y=6 y=5
List all solutions of \sqrt{39y+166}=y+14.