Solve 8^5-7z^2-1=7

$$8 ^ { 5 } - 7 z ^ { 2 } - 1 = 7$$

Solve for z

$$z=6\sqrt{130}\approx 68.410525506$$

$$z=-6\sqrt{130}\approx -68.410525506$$

Steps by Finding Square Root

Calculate 8 to the power of 5 and get 32768.

$$32768-7z^{2}-1=7$$

Subtract 1 from 32768 to get 32767.

$$32767-7z^{2}=7$$

Subtract 32767 from both sides.

$$-7z^{2}=7-32767$$

Subtract 32767 from 7 to get -32760.

$$-7z^{2}=-32760$$

Divide both sides by -7.

$$z^{2}=\frac{-32760}{-7}$$

Divide -32760 by -7 to get 4680.

$$z^{2}=4680$$

Take the square root of both sides of the equation.

$$z=6\sqrt{130} z=-6\sqrt{130}$$

Steps Using the Quadratic Formula

Calculate 8 to the power of 5 and get 32768.

$$32768-7z^{2}-1=7$$

Subtract 1 from 32768 to get 32767.

$$32767-7z^{2}=7$$

Subtract 7 from both sides.

$$32767-7z^{2}-7=0$$

Subtract 7 from 32767 to get 32760.

$$32760-7z^{2}=0$$

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.

$$-7z^{2}+32760=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -7 for a, 0 for b, and 32760 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$z=\frac{0±\sqrt{0^{2}-4\left(-7\right)\times 32760}}{2\left(-7\right)}$$

Square 0.

$$z=\frac{0±\sqrt{-4\left(-7\right)\times 32760}}{2\left(-7\right)}$$

Multiply -4 times -7.

$$z=\frac{0±\sqrt{28\times 32760}}{2\left(-7\right)}$$

Multiply 28 times 32760.

$$z=\frac{0±\sqrt{917280}}{2\left(-7\right)}$$

Take the square root of 917280.

$$z=\frac{0±84\sqrt{130}}{2\left(-7\right)}$$

Multiply 2 times -7.

$$z=\frac{0±84\sqrt{130}}{-14}$$

Now solve the equation z=\frac{0±84\sqrt{130}}{-14} when ± is plus.

$$z=-6\sqrt{130}$$

Now solve the equation z=\frac{0±84\sqrt{130}}{-14} when ± is minus.

$$z=6\sqrt{130}$$

The equation is now solved.

$$z=-6\sqrt{130} z=6\sqrt{130}$$