Solve sqrt{(-5)^{2}+(-6)^{2}+(t-2)^2}=sqrt{(-4)^2+(-3)^2+(t-4)^2}

$$\sqrt { ( - 5 ) ^ { 2 } + ( - 6 ) ^ { 2 } + ( t - 2 ) ^ { 2 } } = \sqrt { ( - 4 ) ^ { 2 } + ( - 3 ) ^ { 2 } + ( t - 4 ) ^ { 2 } }$$

Solve for t

$$t=-6$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{\left(-5\right)^{2}+\left(-6\right)^{2}+\left(t-2\right)^{2}}\right)^{2}=\left(\sqrt{\left(-4\right)^{2}+\left(-3\right)^{2}+\left(t-4\right)^{2}}\right)^{2}$$

Calculate -5 to the power of 2 and get 25.

$$\left(\sqrt{25+\left(-6\right)^{2}+\left(t-2\right)^{2}}\right)^{2}=\left(\sqrt{\left(-4\right)^{2}+\left(-3\right)^{2}+\left(t-4\right)^{2}}\right)^{2}$$

Calculate -6 to the power of 2 and get 36.

$$\left(\sqrt{25+36+\left(t-2\right)^{2}}\right)^{2}=\left(\sqrt{\left(-4\right)^{2}+\left(-3\right)^{2}+\left(t-4\right)^{2}}\right)^{2}$$

Add 25 and 36 to get 61.

$$\left(\sqrt{61+\left(t-2\right)^{2}}\right)^{2}=\left(\sqrt{\left(-4\right)^{2}+\left(-3\right)^{2}+\left(t-4\right)^{2}}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(t-2\right)^{2}.

$$\left(\sqrt{61+t^{2}-4t+4}\right)^{2}=\left(\sqrt{\left(-4\right)^{2}+\left(-3\right)^{2}+\left(t-4\right)^{2}}\right)^{2}$$

Add 61 and 4 to get 65.

$$\left(\sqrt{65+t^{2}-4t}\right)^{2}=\left(\sqrt{\left(-4\right)^{2}+\left(-3\right)^{2}+\left(t-4\right)^{2}}\right)^{2}$$

Calculate \sqrt{65+t^{2}-4t} to the power of 2 and get 65+t^{2}-4t.

$$65+t^{2}-4t=\left(\sqrt{\left(-4\right)^{2}+\left(-3\right)^{2}+\left(t-4\right)^{2}}\right)^{2}$$

Calculate -4 to the power of 2 and get 16.

$$65+t^{2}-4t=\left(\sqrt{16+\left(-3\right)^{2}+\left(t-4\right)^{2}}\right)^{2}$$

Calculate -3 to the power of 2 and get 9.

$$65+t^{2}-4t=\left(\sqrt{16+9+\left(t-4\right)^{2}}\right)^{2}$$

Add 16 and 9 to get 25.

$$65+t^{2}-4t=\left(\sqrt{25+\left(t-4\right)^{2}}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(t-4\right)^{2}.

$$65+t^{2}-4t=\left(\sqrt{25+t^{2}-8t+16}\right)^{2}$$

Add 25 and 16 to get 41.

$$65+t^{2}-4t=\left(\sqrt{41+t^{2}-8t}\right)^{2}$$

Calculate \sqrt{41+t^{2}-8t} to the power of 2 and get 41+t^{2}-8t.

$$65+t^{2}-4t=41+t^{2}-8t$$

Subtract t^{2} from both sides.

$$65+t^{2}-4t-t^{2}=41-8t$$

Combine t^{2} and -t^{2} to get 0.

$$65-4t=41-8t$$

Add 8t to both sides.

$$65-4t+8t=41$$

Combine -4t and 8t to get 4t.

$$65+4t=41$$

Subtract 65 from both sides.

$$4t=41-65$$

Subtract 65 from 41 to get -24.

$$4t=-24$$

Divide both sides by 4.

$$t=\frac{-24}{4}$$

Divide -24 by 4 to get -6.

$$t=-6$$

Substitute -6 for t in the equation \sqrt{\left(-5\right)^{2}+\left(-6\right)^{2}+\left(t-2\right)^{2}}=\sqrt{\left(-4\right)^{2}+\left(-3\right)^{2}+\left(t-4\right)^{2}}.

$$\sqrt{\left(-5\right)^{2}+\left(-6\right)^{2}+\left(-6-2\right)^{2}}=\sqrt{\left(-4\right)^{2}+\left(-3\right)^{2}+\left(-6-4\right)^{2}}$$

Simplify. The value t=-6 satisfies the equation.

$$5\times 5^{\frac{1}{2}}=5\times 5^{\frac{1}{2}}$$

Equation \sqrt{\left(t-2\right)^{2}+61}=\sqrt{\left(t-4\right)^{2}+25} has a unique solution.

$$t=-6$$