Solve (8-2t)^2+(6t)^2=16

$$( 8 - 2 t ) ^ { 2 } + ( 6 t ) ^ { 2 } = 16$$

Solve for t

$$t=\frac{2+\sqrt{26}i}{5}\approx 0.4+1.019803903i$$

$$t=\frac{-\sqrt{26}i+2}{5}\approx 0.4-1.019803903i$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(8-2t\right)^{2}.

$$64-32t+4t^{2}+\left(6t\right)^{2}=16$$

Expand \left(6t\right)^{2}.

$$64-32t+4t^{2}+6^{2}t^{2}=16$$

Calculate 6 to the power of 2 and get 36.

$$64-32t+4t^{2}+36t^{2}=16$$

Combine 4t^{2} and 36t^{2} to get 40t^{2}.

$$64-32t+40t^{2}=16$$

Subtract 16 from both sides.

$$64-32t+40t^{2}-16=0$$

Subtract 16 from 64 to get 48.

$$48-32t+40t^{2}=0$$

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.

$$40t^{2}-32t+48=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 40 for a, -32 for b, and 48 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-32\right)±\sqrt{\left(-32\right)^{2}-4\times 40\times 48}}{2\times 40}$$

Square -32.

$$t=\frac{-\left(-32\right)±\sqrt{1024-4\times 40\times 48}}{2\times 40}$$

Multiply -4 times 40.

$$t=\frac{-\left(-32\right)±\sqrt{1024-160\times 48}}{2\times 40}$$

Multiply -160 times 48.

$$t=\frac{-\left(-32\right)±\sqrt{1024-7680}}{2\times 40}$$

Add 1024 to -7680.

$$t=\frac{-\left(-32\right)±\sqrt{-6656}}{2\times 40}$$

Take the square root of -6656.

$$t=\frac{-\left(-32\right)±16\sqrt{26}i}{2\times 40}$$

The opposite of -32 is 32.

$$t=\frac{32±16\sqrt{26}i}{2\times 40}$$

Multiply 2 times 40.

$$t=\frac{32±16\sqrt{26}i}{80}$$

Now solve the equation t=\frac{32±16\sqrt{26}i}{80} when ± is plus. Add 32 to 16i\sqrt{26}.

$$t=\frac{32+16\sqrt{26}i}{80}$$

Divide 32+16i\sqrt{26} by 80.

$$t=\frac{2+\sqrt{26}i}{5}$$

Now solve the equation t=\frac{32±16\sqrt{26}i}{80} when ± is minus. Subtract 16i\sqrt{26} from 32.

$$t=\frac{-16\sqrt{26}i+32}{80}$$

Divide 32-16i\sqrt{26} by 80.

$$t=\frac{-\sqrt{26}i+2}{5}$$

The equation is now solved.

$$t=\frac{2+\sqrt{26}i}{5} t=\frac{-\sqrt{26}i+2}{5}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(8-2t\right)^{2}.

$$64-32t+4t^{2}+\left(6t\right)^{2}=16$$

Expand \left(6t\right)^{2}.

$$64-32t+4t^{2}+6^{2}t^{2}=16$$

Calculate 6 to the power of 2 and get 36.

$$64-32t+4t^{2}+36t^{2}=16$$

Combine 4t^{2} and 36t^{2} to get 40t^{2}.

$$64-32t+40t^{2}=16$$

Subtract 64 from both sides.

$$-32t+40t^{2}=16-64$$

Subtract 64 from 16 to get -48.

$$-32t+40t^{2}=-48$$

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.

$$40t^{2}-32t=-48$$

Divide both sides by 40.

$$\frac{40t^{2}-32t}{40}=-\frac{48}{40}$$

Dividing by 40 undoes the multiplication by 40.

$$t^{2}+\left(-\frac{32}{40}\right)t=-\frac{48}{40}$$

Reduce the fraction \frac{-32}{40} to lowest terms by extracting and canceling out 8.

$$t^{2}-\frac{4}{5}t=-\frac{48}{40}$$

Reduce the fraction \frac{-48}{40} to lowest terms by extracting and canceling out 8.

$$t^{2}-\frac{4}{5}t=-\frac{6}{5}$$

Divide -\frac{4}{5}, the coefficient of the x term, by 2 to get -\frac{2}{5}. Then add the square of -\frac{2}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{2}{5} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{4}{5}t+\frac{4}{25}=-\frac{6}{5}+\frac{4}{25}$$

Add -\frac{6}{5} to \frac{4}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}-\frac{4}{5}t+\frac{4}{25}=-\frac{26}{25}$$

Factor t^{2}-\frac{4}{5}t+\frac{4}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-\frac{2}{5}\right)^{2}=-\frac{26}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{2}{5}\right)^{2}}=\sqrt{-\frac{26}{25}}$$

Simplify.

$$t-\frac{2}{5}=\frac{\sqrt{26}i}{5} t-\frac{2}{5}=-\frac{\sqrt{26}i}{5}$$

Add \frac{2}{5} to both sides of the equation.

$$t=\frac{2+\sqrt{26}i}{5} t=\frac{-\sqrt{26}i+2}{5}$$