Solve (3sqrt{3})^2=h^2+(frac{3sqrt{3}}{2})^2

$$( 3 \sqrt { 3 } ) ^ { 2 } = h ^ { 2 } + ( \frac { 3 \sqrt { 3 } } { 2 } ) ^ { 2 }$$

Solve for h

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

$$h = -\frac{9}{2} = -4\frac{1}{2} = -4.5$$

Steps Using Difference of Squares

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

$$3^{2}\left(\sqrt{3}\right)^{2}=h^{2}+\left(\frac{3\sqrt{3}}{2}\right)^{2}$$

Calculate 3 to the power of 2 and get 9.

$$9\left(\sqrt{3}\right)^{2}=h^{2}+\left(\frac{3\sqrt{3}}{2}\right)^{2}$$

The square of \sqrt{3} is 3.

$$9\times 3=h^{2}+\left(\frac{3\sqrt{3}}{2}\right)^{2}$$

Multiply 9 and 3 to get 27.

$$27=h^{2}+\left(\frac{3\sqrt{3}}{2}\right)^{2}$$

To raise \frac{3\sqrt{3}}{2} to a power, raise both numerator and denominator to the power and then divide.

$$27=h^{2}+\frac{\left(3\sqrt{3}\right)^{2}}{2^{2}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply h^{2} times \frac{2^{2}}{2^{2}}.

$$27=\frac{h^{2}\times 2^{2}}{2^{2}}+\frac{\left(3\sqrt{3}\right)^{2}}{2^{2}}$$

Since \frac{h^{2}\times 2^{2}}{2^{2}} and \frac{\left(3\sqrt{3}\right)^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$27=\frac{h^{2}\times 2^{2}+\left(3\sqrt{3}\right)^{2}}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$27=\frac{h^{2}\times 4+\left(3\sqrt{3}\right)^{2}}{2^{2}}$$

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

$$27=\frac{h^{2}\times 4+3^{2}\left(\sqrt{3}\right)^{2}}{2^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$27=\frac{h^{2}\times 4+9\left(\sqrt{3}\right)^{2}}{2^{2}}$$

The square of \sqrt{3} is 3.

$$27=\frac{h^{2}\times 4+9\times 3}{2^{2}}$$

Multiply 9 and 3 to get 27.

$$27=\frac{h^{2}\times 4+27}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$27=\frac{h^{2}\times 4+27}{4}$$

Divide each term of h^{2}\times 4+27 by 4 to get h^{2}+\frac{27}{4}.

$$27=h^{2}+\frac{27}{4}$$

Swap sides so that all variable terms are on the left hand side.

$$h^{2}+\frac{27}{4}=27$$

Subtract 27 from both sides.

$$h^{2}+\frac{27}{4}-27=0$$

Subtract 27 from \frac{27}{4} to get -\frac{81}{4}.

$$h^{2}-\frac{81}{4}=0$$

Multiply both sides by 4.

$$4h^{2}-81=0$$

Consider 4h^{2}-81. Rewrite 4h^{2}-81 as \left(2h\right)^{2}-9^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(2h-9\right)\left(2h+9\right)=0$$

To find equation solutions, solve 2h-9=0 and 2h+9=0.

$$h=\frac{9}{2} h=-\frac{9}{2}$$

Steps by Finding Square Root

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

$$3^{2}\left(\sqrt{3}\right)^{2}=h^{2}+\left(\frac{3\sqrt{3}}{2}\right)^{2}$$

Calculate 3 to the power of 2 and get 9.

$$9\left(\sqrt{3}\right)^{2}=h^{2}+\left(\frac{3\sqrt{3}}{2}\right)^{2}$$

The square of \sqrt{3} is 3.

$$9\times 3=h^{2}+\left(\frac{3\sqrt{3}}{2}\right)^{2}$$

Multiply 9 and 3 to get 27.

$$27=h^{2}+\left(\frac{3\sqrt{3}}{2}\right)^{2}$$

To raise \frac{3\sqrt{3}}{2} to a power, raise both numerator and denominator to the power and then divide.

$$27=h^{2}+\frac{\left(3\sqrt{3}\right)^{2}}{2^{2}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply h^{2} times \frac{2^{2}}{2^{2}}.

$$27=\frac{h^{2}\times 2^{2}}{2^{2}}+\frac{\left(3\sqrt{3}\right)^{2}}{2^{2}}$$

Since \frac{h^{2}\times 2^{2}}{2^{2}} and \frac{\left(3\sqrt{3}\right)^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$27=\frac{h^{2}\times 2^{2}+\left(3\sqrt{3}\right)^{2}}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$27=\frac{h^{2}\times 4+\left(3\sqrt{3}\right)^{2}}{2^{2}}$$

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

$$27=\frac{h^{2}\times 4+3^{2}\left(\sqrt{3}\right)^{2}}{2^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$27=\frac{h^{2}\times 4+9\left(\sqrt{3}\right)^{2}}{2^{2}}$$

The square of \sqrt{3} is 3.

$$27=\frac{h^{2}\times 4+9\times 3}{2^{2}}$$

Multiply 9 and 3 to get 27.

$$27=\frac{h^{2}\times 4+27}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$27=\frac{h^{2}\times 4+27}{4}$$

Divide each term of h^{2}\times 4+27 by 4 to get h^{2}+\frac{27}{4}.

$$27=h^{2}+\frac{27}{4}$$

Swap sides so that all variable terms are on the left hand side.

$$h^{2}+\frac{27}{4}=27$$

Subtract \frac{27}{4} from both sides.

$$h^{2}=27-\frac{27}{4}$$

Subtract \frac{27}{4} from 27 to get \frac{81}{4}.

$$h^{2}=\frac{81}{4}$$

Take the square root of both sides of the equation.

$$h=\frac{9}{2} h=-\frac{9}{2}$$

Steps Using the Quadratic Formula

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

$$3^{2}\left(\sqrt{3}\right)^{2}=h^{2}+\left(\frac{3\sqrt{3}}{2}\right)^{2}$$

Calculate 3 to the power of 2 and get 9.

$$9\left(\sqrt{3}\right)^{2}=h^{2}+\left(\frac{3\sqrt{3}}{2}\right)^{2}$$

The square of \sqrt{3} is 3.

$$9\times 3=h^{2}+\left(\frac{3\sqrt{3}}{2}\right)^{2}$$

Multiply 9 and 3 to get 27.

$$27=h^{2}+\left(\frac{3\sqrt{3}}{2}\right)^{2}$$

To raise \frac{3\sqrt{3}}{2} to a power, raise both numerator and denominator to the power and then divide.

$$27=h^{2}+\frac{\left(3\sqrt{3}\right)^{2}}{2^{2}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply h^{2} times \frac{2^{2}}{2^{2}}.

$$27=\frac{h^{2}\times 2^{2}}{2^{2}}+\frac{\left(3\sqrt{3}\right)^{2}}{2^{2}}$$

Since \frac{h^{2}\times 2^{2}}{2^{2}} and \frac{\left(3\sqrt{3}\right)^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$27=\frac{h^{2}\times 2^{2}+\left(3\sqrt{3}\right)^{2}}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$27=\frac{h^{2}\times 4+\left(3\sqrt{3}\right)^{2}}{2^{2}}$$

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

$$27=\frac{h^{2}\times 4+3^{2}\left(\sqrt{3}\right)^{2}}{2^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$27=\frac{h^{2}\times 4+9\left(\sqrt{3}\right)^{2}}{2^{2}}$$

The square of \sqrt{3} is 3.

$$27=\frac{h^{2}\times 4+9\times 3}{2^{2}}$$

Multiply 9 and 3 to get 27.

$$27=\frac{h^{2}\times 4+27}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$27=\frac{h^{2}\times 4+27}{4}$$

Divide each term of h^{2}\times 4+27 by 4 to get h^{2}+\frac{27}{4}.

$$27=h^{2}+\frac{27}{4}$$

Swap sides so that all variable terms are on the left hand side.

$$h^{2}+\frac{27}{4}=27$$

Subtract 27 from both sides.

$$h^{2}+\frac{27}{4}-27=0$$

Subtract 27 from \frac{27}{4} to get -\frac{81}{4}.

$$h^{2}-\frac{81}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -\frac{81}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$h=\frac{0±\sqrt{0^{2}-4\left(-\frac{81}{4}\right)}}{2}$$

Square 0.

$$h=\frac{0±\sqrt{-4\left(-\frac{81}{4}\right)}}{2}$$

Multiply -4 times -\frac{81}{4}.

$$h=\frac{0±\sqrt{81}}{2}$$

Take the square root of 81.

$$h=\frac{0±9}{2}$$

Now solve the equation h=\frac{0±9}{2} when ± is plus. Divide 9 by 2.

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

Now solve the equation h=\frac{0±9}{2} when ± is minus. Divide -9 by 2.

$$h=-\frac{9}{2}$$

The equation is now solved.

$$h=\frac{9}{2} h=-\frac{9}{2}$$