Solve sqrt{25}-sqrt[3]{8}-sqrt{121}+sqrt{49/16}

$$\sqrt { 25 } - \sqrt[ 3 ] { 8 } - \sqrt { 121 } + \sqrt { \frac { 49 } { 16 } }$$

Evaluate

$$-\frac{25}{4}=-6.25$$

Solution Steps

Calculate the square root of 25 and get 5.

$$5-\sqrt[3]{8}-\sqrt{121}+\sqrt{\frac{49}{16}}$$

Calculate \sqrt[3]{8} and get 2.

$$5-2-\sqrt{121}+\sqrt{\frac{49}{16}}$$

Subtract 2 from 5 to get 3.

$$3-\sqrt{121}+\sqrt{\frac{49}{16}}$$

Calculate the square root of 121 and get 11.

$$3-11+\sqrt{\frac{49}{16}}$$

Subtract 11 from 3 to get -8.

$$-8+\sqrt{\frac{49}{16}}$$

Rewrite the square root of the division \frac{49}{16} as the division of square roots \frac{\sqrt{49}}{\sqrt{16}}. Take the square root of both numerator and denominator.

$$-8+\frac{7}{4}$$

Add -8 and \frac{7}{4} to get -\frac{25}{4}.

$$-\frac{25}{4}$$

Factor

$$-\frac{25}{4} = -6\frac{1}{4} = -6.25$$