Solve sqrt{4}+2+(2+5+2*5)-19+3*(5^2-1-4^2)=

$$\sqrt { 4 } + 2 + ( 2 + 5 + 2 \cdot 5 ) - 19 + 3 \cdot ( 5 ^ { 2 } - 1 - 4 ^ { 2 } ) =$$

Evaluate

$$26$$

Solution Steps

Calculate the square root of 4 and get 2.

$$2+2+2+5+2\times 5-19+3\left(5^{2}-1-4^{2}\right)$$

Add 2 and 2 to get 4.

$$4+2+5+2\times 5-19+3\left(5^{2}-1-4^{2}\right)$$

Add 4 and 2 to get 6.

$$6+5+2\times 5-19+3\left(5^{2}-1-4^{2}\right)$$

Add 6 and 5 to get 11.

$$11+2\times 5-19+3\left(5^{2}-1-4^{2}\right)$$

Multiply 2 and 5 to get 10.

$$11+10-19+3\left(5^{2}-1-4^{2}\right)$$

Add 11 and 10 to get 21.

$$21-19+3\left(5^{2}-1-4^{2}\right)$$

Subtract 19 from 21 to get 2.

$$2+3\left(5^{2}-1-4^{2}\right)$$

Calculate 5 to the power of 2 and get 25.

$$2+3\left(25-1-4^{2}\right)$$

Subtract 1 from 25 to get 24.

$$2+3\left(24-4^{2}\right)$$

Calculate 4 to the power of 2 and get 16.

$$2+3\left(24-16\right)$$

Subtract 16 from 24 to get 8.

$$2+3\times 8$$

Multiply 3 and 8 to get 24.

$$2+24$$

Add 2 and 24 to get 26.

$$26$$

Factor

$$2\times 13$$