Solve (3-t)^2+(4-t)^2+1+t^2=16

$$( 3 - t ) ^ { 2 } + ( 4 - t ) ^ { 2 } + 1 + t ^ { 2 } = 16$$

Solve for t

$$t = \frac{\sqrt{19} + 7}{3} \approx 3.786299648$$

$$t=\frac{7-\sqrt{19}}{3}\approx 0.880367019$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3-t\right)^{2}.

$$9-6t+t^{2}+\left(4-t\right)^{2}+1+t^{2}=16$$

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

$$9-6t+t^{2}+16-8t+t^{2}+1+t^{2}=16$$

Add 9 and 16 to get 25.

$$25-6t+t^{2}-8t+t^{2}+1+t^{2}=16$$

Combine -6t and -8t to get -14t.

$$25-14t+t^{2}+t^{2}+1+t^{2}=16$$

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

$$25-14t+2t^{2}+1+t^{2}=16$$

Add 25 and 1 to get 26.

$$26-14t+2t^{2}+t^{2}=16$$

Combine 2t^{2} and t^{2} to get 3t^{2}.

$$26-14t+3t^{2}=16$$

Subtract 16 from both sides.

$$26-14t+3t^{2}-16=0$$

Subtract 16 from 26 to get 10.

$$10-14t+3t^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$3t^{2}-14t+10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -14 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 3\times 10}}{2\times 3}$$

Square -14.

$$t=\frac{-\left(-14\right)±\sqrt{196-4\times 3\times 10}}{2\times 3}$$

Multiply -4 times 3.

$$t=\frac{-\left(-14\right)±\sqrt{196-12\times 10}}{2\times 3}$$

Multiply -12 times 10.

$$t=\frac{-\left(-14\right)±\sqrt{196-120}}{2\times 3}$$

Add 196 to -120.

$$t=\frac{-\left(-14\right)±\sqrt{76}}{2\times 3}$$

Take the square root of 76.

$$t=\frac{-\left(-14\right)±2\sqrt{19}}{2\times 3}$$

The opposite of -14 is 14.

$$t=\frac{14±2\sqrt{19}}{2\times 3}$$

Multiply 2 times 3.

$$t=\frac{14±2\sqrt{19}}{6}$$

Now solve the equation t=\frac{14±2\sqrt{19}}{6} when ± is plus. Add 14 to 2\sqrt{19}.

$$t=\frac{2\sqrt{19}+14}{6}$$

Divide 14+2\sqrt{19} by 6.

$$t=\frac{\sqrt{19}+7}{3}$$

Now solve the equation t=\frac{14±2\sqrt{19}}{6} when ± is minus. Subtract 2\sqrt{19} from 14.

$$t=\frac{14-2\sqrt{19}}{6}$$

Divide 14-2\sqrt{19} by 6.

$$t=\frac{7-\sqrt{19}}{3}$$

The equation is now solved.

$$t=\frac{\sqrt{19}+7}{3} t=\frac{7-\sqrt{19}}{3}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3-t\right)^{2}.

$$9-6t+t^{2}+\left(4-t\right)^{2}+1+t^{2}=16$$

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

$$9-6t+t^{2}+16-8t+t^{2}+1+t^{2}=16$$

Add 9 and 16 to get 25.

$$25-6t+t^{2}-8t+t^{2}+1+t^{2}=16$$

Combine -6t and -8t to get -14t.

$$25-14t+t^{2}+t^{2}+1+t^{2}=16$$

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

$$25-14t+2t^{2}+1+t^{2}=16$$

Add 25 and 1 to get 26.

$$26-14t+2t^{2}+t^{2}=16$$

Combine 2t^{2} and t^{2} to get 3t^{2}.

$$26-14t+3t^{2}=16$$

Subtract 26 from both sides.

$$-14t+3t^{2}=16-26$$

Subtract 26 from 16 to get -10.

$$-14t+3t^{2}=-10$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$3t^{2}-14t=-10$$

Divide both sides by 3.

$$\frac{3t^{2}-14t}{3}=-\frac{10}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$t^{2}-\frac{14}{3}t=-\frac{10}{3}$$

Divide -\frac{14}{3}, the coefficient of the x term, by 2 to get -\frac{7}{3}. Then add the square of -\frac{7}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-\frac{14}{3}t+\left(-\frac{7}{3}\right)^{2}=-\frac{10}{3}+\left(-\frac{7}{3}\right)^{2}$$

Square -\frac{7}{3} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{14}{3}t+\frac{49}{9}=-\frac{10}{3}+\frac{49}{9}$$

Add -\frac{10}{3} to \frac{49}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}-\frac{14}{3}t+\frac{49}{9}=\frac{19}{9}$$

Factor t^{2}-\frac{14}{3}t+\frac{49}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-\frac{7}{3}\right)^{2}=\frac{19}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{7}{3}\right)^{2}}=\sqrt{\frac{19}{9}}$$

Simplify.

$$t-\frac{7}{3}=\frac{\sqrt{19}}{3} t-\frac{7}{3}=-\frac{\sqrt{19}}{3}$$

Add \frac{7}{3} to both sides of the equation.

$$t=\frac{\sqrt{19}+7}{3} t=\frac{7-\sqrt{19}}{3}$$