Solve 301^2-299^2/(1,3)^2+(1,3)*2(2,7)+(2,7)^2

$$\frac { 301 ^ { 2 } - 299 ^ { 2 } } { ( 1,3 ) ^ { 2 } + ( 1,3 ) \cdot 2 ( 2,7 ) + ( 2,7 ) ^ { 2 } }$$

Evaluate

$$75$$

Solution Steps

Calculate 301 to the power of 2 and get 90601.

$$\frac{90601-299^{2}}{1,3^{2}+1,3\times 2\times 2,7+2,7^{2}}$$

Calculate 299 to the power of 2 and get 89401.

$$\frac{90601-89401}{1,3^{2}+1,3\times 2\times 2,7+2,7^{2}}$$

Subtract 89401 from 90601 to get 1200.

$$\frac{1200}{1,3^{2}+1,3\times 2\times 2,7+2,7^{2}}$$

Calculate 1,3 to the power of 2 and get 1,69.

$$\frac{1200}{1,69+1,3\times 2\times 2,7+2,7^{2}}$$

Multiply 1,3 and 2 to get 2,6.

$$\frac{1200}{1,69+2,6\times 2,7+2,7^{2}}$$

Multiply 2,6 and 2,7 to get 7,02.

$$\frac{1200}{1,69+7,02+2,7^{2}}$$

Add 1,69 and 7,02 to get 8,71.

$$\frac{1200}{8,71+2,7^{2}}$$

Calculate 2,7 to the power of 2 and get 7,29.

$$\frac{1200}{8,71+7,29}$$

Add 8,71 and 7,29 to get 16.

$$\frac{1200}{16}$$

Divide 1200 by 16 to get 75.

$$75$$

Factor

$$3\times 5^{2}$$