Solve -648=-8k^2

$$- 648 = - 8 k ^ { 2 }$$

Solve for k

$$k=9$$

$$k=-9$$

Steps Using Difference of Squares

Divide both sides by -8.

$$\frac{-648}{-8}=k^{2}$$

Divide -648 by -8 to get 81.

$$81=k^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$k^{2}=81$$

Subtract 81 from both sides.

$$k^{2}-81=0$$

Consider k^{2}-81. Rewrite k^{2}-81 as k^{2}-9^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(k-9\right)\left(k+9\right)=0$$

To find equation solutions, solve k-9=0 and k+9=0.

$$k=9 k=-9$$

Steps by Finding Square Root

Divide both sides by -8.

$$\frac{-648}{-8}=k^{2}$$

Divide -648 by -8 to get 81.

$$81=k^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$k^{2}=81$$

Take the square root of both sides of the equation.

$$k=9 k=-9$$

Steps Using the Quadratic Formula

Divide both sides by -8.

$$\frac{-648}{-8}=k^{2}$$

Divide -648 by -8 to get 81.

$$81=k^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$k^{2}=81$$

Subtract 81 from both sides.

$$k^{2}-81=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -81 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{0±\sqrt{0^{2}-4\left(-81\right)}}{2}$$

Square 0.

$$k=\frac{0±\sqrt{-4\left(-81\right)}}{2}$$

Multiply -4 times -81.

$$k=\frac{0±\sqrt{324}}{2}$$

Take the square root of 324.

$$k=\frac{0±18}{2}$$

Now solve the equation k=\frac{0±18}{2} when ± is plus. Divide 18 by 2.

$$k=9$$

Now solve the equation k=\frac{0±18}{2} when ± is minus. Divide -18 by 2.

$$k=-9$$

The equation is now solved.

$$k=9 k=-9$$