Solve 9^2/20^2+y^2/16^2=1

$$\frac{ { 9 }^{ 2 } }{ { 20 }^{ 2 } } + \frac{ { y }^{ 2 } }{ { 16 }^{ 2 } } =1$$

Solve for y

$$y = \frac{4 \sqrt{319}}{5} \approx 14.28845688$$

$$y = -\frac{4 \sqrt{319}}{5} \approx -14.28845688$$

Steps by Finding Square Root

Calculate 9 to the power of 2 and get 81.

$$\frac{81}{20^{2}}+\frac{y^{2}}{16^{2}}=1$$

Calculate 20 to the power of 2 and get 400.

$$\frac{81}{400}+\frac{y^{2}}{16^{2}}=1$$

Calculate 16 to the power of 2 and get 256.

$$\frac{81}{400}+\frac{y^{2}}{256}=1$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 400 and 256 is 6400. Multiply \frac{81}{400} times \frac{16}{16}. Multiply \frac{y^{2}}{256} times \frac{25}{25}.

$$\frac{81\times 16}{6400}+\frac{25y^{2}}{6400}=1$$

Since \frac{81\times 16}{6400} and \frac{25y^{2}}{6400} have the same denominator, add them by adding their numerators.

$$\frac{81\times 16+25y^{2}}{6400}=1$$

Do the multiplications in 81\times 16+25y^{2}.

$$\frac{1296+25y^{2}}{6400}=1$$

Divide each term of 1296+25y^{2} by 6400 to get \frac{81}{400}+\frac{1}{256}y^{2}.

$$\frac{81}{400}+\frac{1}{256}y^{2}=1$$

Subtract \frac{81}{400} from both sides.

$$\frac{1}{256}y^{2}=1-\frac{81}{400}$$

Subtract \frac{81}{400} from 1 to get \frac{319}{400}.

$$\frac{1}{256}y^{2}=\frac{319}{400}$$

Multiply both sides by 256, the reciprocal of \frac{1}{256}.

$$y^{2}=\frac{319}{400}\times 256$$

Multiply \frac{319}{400} and 256 to get \frac{5104}{25}.

$$y^{2}=\frac{5104}{25}$$

Take the square root of both sides of the equation.

$$y=\frac{4\sqrt{319}}{5} y=-\frac{4\sqrt{319}}{5}$$

Steps Using the Quadratic Formula

Calculate 9 to the power of 2 and get 81.

$$\frac{81}{20^{2}}+\frac{y^{2}}{16^{2}}=1$$

Calculate 20 to the power of 2 and get 400.

$$\frac{81}{400}+\frac{y^{2}}{16^{2}}=1$$

Calculate 16 to the power of 2 and get 256.

$$\frac{81}{400}+\frac{y^{2}}{256}=1$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 400 and 256 is 6400. Multiply \frac{81}{400} times \frac{16}{16}. Multiply \frac{y^{2}}{256} times \frac{25}{25}.

$$\frac{81\times 16}{6400}+\frac{25y^{2}}{6400}=1$$

Since \frac{81\times 16}{6400} and \frac{25y^{2}}{6400} have the same denominator, add them by adding their numerators.

$$\frac{81\times 16+25y^{2}}{6400}=1$$

Do the multiplications in 81\times 16+25y^{2}.

$$\frac{1296+25y^{2}}{6400}=1$$

Divide each term of 1296+25y^{2} by 6400 to get \frac{81}{400}+\frac{1}{256}y^{2}.

$$\frac{81}{400}+\frac{1}{256}y^{2}=1$$

Subtract 1 from both sides.

$$\frac{81}{400}+\frac{1}{256}y^{2}-1=0$$

Subtract 1 from \frac{81}{400} to get -\frac{319}{400}.

$$-\frac{319}{400}+\frac{1}{256}y^{2}=0$$

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$\frac{1}{256}y^{2}-\frac{319}{400}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{256} for a, 0 for b, and -\frac{319}{400} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{0±\sqrt{0^{2}-4\times \frac{1}{256}\left(-\frac{319}{400}\right)}}{2\times \frac{1}{256}}$$

Square 0.

$$y=\frac{0±\sqrt{-4\times \frac{1}{256}\left(-\frac{319}{400}\right)}}{2\times \frac{1}{256}}$$

Multiply -4 times \frac{1}{256}.

$$y=\frac{0±\sqrt{-\frac{1}{64}\left(-\frac{319}{400}\right)}}{2\times \frac{1}{256}}$$

Multiply -\frac{1}{64} times -\frac{319}{400} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$y=\frac{0±\sqrt{\frac{319}{25600}}}{2\times \frac{1}{256}}$$

Take the square root of \frac{319}{25600}.

$$y=\frac{0±\frac{\sqrt{319}}{160}}{2\times \frac{1}{256}}$$

Multiply 2 times \frac{1}{256}.

$$y=\frac{0±\frac{\sqrt{319}}{160}}{\frac{1}{128}}$$

Now solve the equation y=\frac{0±\frac{\sqrt{319}}{160}}{\frac{1}{128}} when ± is plus.

$$y=\frac{4\sqrt{319}}{5}$$

Now solve the equation y=\frac{0±\frac{\sqrt{319}}{160}}{\frac{1}{128}} when ± is minus.

$$y=-\frac{4\sqrt{319}}{5}$$

The equation is now solved.

$$y=\frac{4\sqrt{319}}{5} y=-\frac{4\sqrt{319}}{5}$$