Solve 3k^2-8=-5

$$3 k ^ { 2 } - 8 = - 5$$

Solve for k

$$k=1$$

$$k=-1$$

Steps Using Difference of Squares

Add 5 to both sides.

$$3k^{2}-8+5=0$$

Add -8 and 5 to get -3.

$$3k^{2}-3=0$$

Divide both sides by 3.

$$k^{2}-1=0$$

Consider k^{2}-1. Rewrite k^{2}-1 as k^{2}-1^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(k-1\right)\left(k+1\right)=0$$

To find equation solutions, solve k-1=0 and k+1=0.

$$k=1 k=-1$$

Steps by Finding Square Root

Add 8 to both sides.

$$3k^{2}=-5+8$$

Add -5 and 8 to get 3.

$$3k^{2}=3$$

Divide both sides by 3.

$$k^{2}=\frac{3}{3}$$

Divide 3 by 3 to get 1.

$$k^{2}=1$$

Take the square root of both sides of the equation.

$$k=1 k=-1$$

Steps Using the Quadratic Formula

Add 5 to both sides.

$$3k^{2}-8+5=0$$

Add -8 and 5 to get -3.

$$3k^{2}-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 0 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{0±\sqrt{0^{2}-4\times 3\left(-3\right)}}{2\times 3}$$

Square 0.

$$k=\frac{0±\sqrt{-4\times 3\left(-3\right)}}{2\times 3}$$

Multiply -4 times 3.

$$k=\frac{0±\sqrt{-12\left(-3\right)}}{2\times 3}$$

Multiply -12 times -3.

$$k=\frac{0±\sqrt{36}}{2\times 3}$$

Take the square root of 36.

$$k=\frac{0±6}{2\times 3}$$

Multiply 2 times 3.

$$k=\frac{0±6}{6}$$

Now solve the equation k=\frac{0±6}{6} when ± is plus. Divide 6 by 6.

$$k=1$$

Now solve the equation k=\frac{0±6}{6} when ± is minus. Divide -6 by 6.

$$k=-1$$

The equation is now solved.

$$k=1 k=-1$$