Solve left(4-3right)+left(1+2right)^2/6+left(7-5right)

$$\frac{ \left( 4-3 \right) + { \left( 1+2 \right) }^{ 2 } }{ 6+ \left( 7-5 \right) }$$

Evaluate

$$\frac{5}{4}=1.25$$

Solution Steps

Subtract 3 from 4 to get 1.

$$\frac{1+\left(1+2\right)^{2}}{6+7-5}$$

Add 1 and 2 to get 3.

$$\frac{1+3^{2}}{6+7-5}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{1+9}{6+7-5}$$

Add 1 and 9 to get 10.

$$\frac{10}{6+7-5}$$

Add 6 and 7 to get 13.

$$\frac{10}{13-5}$$

Subtract 5 from 13 to get 8.

$$\frac{10}{8}$$

Reduce the fraction \frac{10}{8} to lowest terms by extracting and canceling out 2.

$$\frac{5}{4}$$

Factor

$$\frac{5}{2 ^ {2}} = 1\frac{1}{4} = 1.25$$