Solve 1^3-7d^2+48=0

$$1 ^ { 3 } - 7 d ^ { 2 } + 48 = 0$$

Solve for d

$$d=\sqrt{7}\approx 2.645751311$$

$$d=-\sqrt{7}\approx -2.645751311$$

Steps by Finding Square Root

Calculate 1 to the power of 3 and get 1.

$$1-7d^{2}+48=0$$

Add 1 and 48 to get 49.

$$49-7d^{2}=0$$

Subtract 49 from both sides. Anything subtracted from zero gives its negation.

$$-7d^{2}=-49$$

Divide both sides by -7.

$$d^{2}=\frac{-49}{-7}$$

Divide -49 by -7 to get 7.

$$d^{2}=7$$

Take the square root of both sides of the equation.

$$d=\sqrt{7} d=-\sqrt{7}$$

Steps Using the Quadratic Formula

Calculate 1 to the power of 3 and get 1.

$$1-7d^{2}+48=0$$

Add 1 and 48 to get 49.

$$49-7d^{2}=0$$

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$-7d^{2}+49=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -7 for a, 0 for b, and 49 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$d=\frac{0±\sqrt{0^{2}-4\left(-7\right)\times 49}}{2\left(-7\right)}$$

Square 0.

$$d=\frac{0±\sqrt{-4\left(-7\right)\times 49}}{2\left(-7\right)}$$

Multiply -4 times -7.

$$d=\frac{0±\sqrt{28\times 49}}{2\left(-7\right)}$$

Multiply 28 times 49.

$$d=\frac{0±\sqrt{1372}}{2\left(-7\right)}$$

Take the square root of 1372.

$$d=\frac{0±14\sqrt{7}}{2\left(-7\right)}$$

Multiply 2 times -7.

$$d=\frac{0±14\sqrt{7}}{-14}$$

Now solve the equation d=\frac{0±14\sqrt{7}}{-14} when ± is plus.

$$d=-\sqrt{7}$$

Now solve the equation d=\frac{0±14\sqrt{7}}{-14} when ± is minus.

$$d=\sqrt{7}$$

The equation is now solved.

$$d=-\sqrt{7} d=\sqrt{7}$$