Solve ((3+sqrt{190})^2+(3+sqrt{4})^2)*5

$$( ( 3 + \sqrt { 190 } ) ^ { 2 } + ( 3 + \sqrt { 4 } ) ^ { 2 } ) * 5$$

Evaluate

$$30\sqrt{190}+1120\approx 1533.521462563$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3+\sqrt{190}\right)^{2}.

$$5\left(9+6\sqrt{190}+\left(\sqrt{190}\right)^{2}+\left(3+\sqrt{4}\right)^{2}\right)$$

The square of \sqrt{190} is 190.

$$5\left(9+6\sqrt{190}+190+\left(3+\sqrt{4}\right)^{2}\right)$$

Add 9 and 190 to get 199.

$$5\left(199+6\sqrt{190}+\left(3+\sqrt{4}\right)^{2}\right)$$

Calculate the square root of 4 and get 2.

$$5\left(199+6\sqrt{190}+\left(3+2\right)^{2}\right)$$

Add 3 and 2 to get 5.

$$5\left(199+6\sqrt{190}+5^{2}\right)$$

Calculate 5 to the power of 2 and get 25.

$$5\left(199+6\sqrt{190}+25\right)$$

Add 199 and 25 to get 224.

$$5\left(224+6\sqrt{190}\right)$$

Use the distributive property to multiply 5 by 224+6\sqrt{190}.

$$1120+30\sqrt{190}$$

Expand

$$30 \sqrt{190} + 1120 = 1533.521462563$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3+\sqrt{190}\right)^{2}.

$$5\left(9+6\sqrt{190}+\left(\sqrt{190}\right)^{2}+\left(3+\sqrt{4}\right)^{2}\right)$$

The square of \sqrt{190} is 190.

$$5\left(9+6\sqrt{190}+190+\left(3+\sqrt{4}\right)^{2}\right)$$

Add 9 and 190 to get 199.

$$5\left(199+6\sqrt{190}+\left(3+\sqrt{4}\right)^{2}\right)$$

Calculate the square root of 4 and get 2.

$$5\left(199+6\sqrt{190}+\left(3+2\right)^{2}\right)$$

Add 3 and 2 to get 5.

$$5\left(199+6\sqrt{190}+5^{2}\right)$$

Calculate 5 to the power of 2 and get 25.

$$5\left(199+6\sqrt{190}+25\right)$$

Add 199 and 25 to get 224.

$$5\left(224+6\sqrt{190}\right)$$

Use the distributive property to multiply 5 by 224+6\sqrt{190}.

$$1120+30\sqrt{190}$$