Solve (7)-6sqrt{45}*(-sqrt{48})

$$( 7 ) - 6 \sqrt { 45 } \times ( - \sqrt { 48 } )$$

Evaluate

$$72\sqrt{15}+7\approx 285.854800927$$

Solution Steps

Factor 45=3^{2}\times 5. Rewrite the square root of the product \sqrt{3^{2}\times 5} as the product of square roots \sqrt{3^{2}}\sqrt{5}. Take the square root of 3^{2}.

$$7-6\times 3\sqrt{5}\left(-\sqrt{48}\right)$$

Multiply 6 and 3 to get 18.

$$7-18\sqrt{5}\left(-\sqrt{48}\right)$$

Factor 48=4^{2}\times 3. Rewrite the square root of the product \sqrt{4^{2}\times 3} as the product of square roots \sqrt{4^{2}}\sqrt{3}. Take the square root of 4^{2}.

$$7-18\sqrt{5}\left(-4\sqrt{3}\right)$$

Multiply -1 and 18 to get -18.

$$7-18\sqrt{5}\left(-1\right)\times 4\sqrt{3}$$

Multiply -18 and -1 to get 18.

$$7+18\sqrt{5}\times 4\sqrt{3}$$

Multiply 18 and 4 to get 72.

$$7+72\sqrt{5}\sqrt{3}$$

To multiply \sqrt{5} and \sqrt{3}, multiply the numbers under the square root.

$$7+72\sqrt{15}$$