Solve [4(10-2)/5]^2+[6-3(10-t)/5]^2=(10-2t)^2

$$[ \frac { 4 ( 10 - 2 ) } { 5 } ] ^ { 2 } + [ 6 - \frac { 3 ( 10 - t ) } { 5 } ] ^ { 2 } = ( 10 - 2 t ) ^ { 2 }$$

Solve for t

$$t = \frac{2 \sqrt{28921} + 500}{91} \approx 9.23212645$$

$$t = \frac{500 - 2 \sqrt{28921}}{91} \approx 1.756884539$$

Steps Using the Quadratic Formula

Subtract 2 from 10 to get 8.

$$\left(\frac{4\times 8}{5}\right)^{2}+\left(6-\frac{3\left(10-t\right)}{5}\right)^{2}=\left(10-2t\right)^{2}$$

Multiply 4 and 8 to get 32.

$$\left(\frac{32}{5}\right)^{2}+\left(6-\frac{3\left(10-t\right)}{5}\right)^{2}=\left(10-2t\right)^{2}$$

Calculate \frac{32}{5} to the power of 2 and get \frac{1024}{25}.

$$\frac{1024}{25}+\left(6-\frac{3\left(10-t\right)}{5}\right)^{2}=\left(10-2t\right)^{2}$$

Use the distributive property to multiply 3 by 10-t.

$$\frac{1024}{25}+\left(6-\frac{30-3t}{5}\right)^{2}=\left(10-2t\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(6-\frac{30-3t}{5}\right)^{2}.

$$\frac{1024}{25}+36+12\left(-\frac{30-3t}{5}\right)+\left(-\frac{30-3t}{5}\right)^{2}=\left(10-2t\right)^{2}$$

Express 12\left(-\frac{30-3t}{5}\right) as a single fraction.

$$\frac{1024}{25}+36+\frac{-12\left(30-3t\right)}{5}+\left(-\frac{30-3t}{5}\right)^{2}=\left(10-2t\right)^{2}$$

Calculate -\frac{30-3t}{5} to the power of 2 and get \left(\frac{30-3t}{5}\right)^{2}.

$$\frac{1024}{25}+36+\frac{-12\left(30-3t\right)}{5}+\left(\frac{30-3t}{5}\right)^{2}=\left(10-2t\right)^{2}$$

Add \frac{1024}{25} and 36 to get \frac{1924}{25}.

$$\frac{1924}{25}+\frac{-12\left(30-3t\right)}{5}+\left(\frac{30-3t}{5}\right)^{2}=\left(10-2t\right)^{2}$$

To raise \frac{30-3t}{5} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{1924}{25}+\frac{-12\left(30-3t\right)}{5}+\frac{\left(30-3t\right)^{2}}{5^{2}}=\left(10-2t\right)^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 25 and 5 is 25. Multiply \frac{-12\left(30-3t\right)}{5} times \frac{5}{5}.

$$\frac{1924}{25}+\frac{5\left(-12\right)\left(30-3t\right)}{25}+\frac{\left(30-3t\right)^{2}}{5^{2}}=\left(10-2t\right)^{2}$$

Since \frac{1924}{25} and \frac{5\left(-12\right)\left(30-3t\right)}{25} have the same denominator, add them by adding their numerators.

$$\frac{1924+5\left(-12\right)\left(30-3t\right)}{25}+\frac{\left(30-3t\right)^{2}}{5^{2}}=\left(10-2t\right)^{2}$$

Do the multiplications in 1924+5\left(-12\right)\left(30-3t\right).

$$\frac{1924-1800+180t}{25}+\frac{\left(30-3t\right)^{2}}{5^{2}}=\left(10-2t\right)^{2}$$

Combine like terms in 1924-1800+180t.

$$\frac{180t+124}{25}+\frac{\left(30-3t\right)^{2}}{5^{2}}=\left(10-2t\right)^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Expand 5^{2}.

$$\frac{180t+124}{25}+\frac{\left(30-3t\right)^{2}}{25}=\left(10-2t\right)^{2}$$

Since \frac{180t+124}{25} and \frac{\left(30-3t\right)^{2}}{25} have the same denominator, add them by adding their numerators.

$$\frac{180t+124+\left(30-3t\right)^{2}}{25}=\left(10-2t\right)^{2}$$

Do the multiplications in 180t+124+\left(30-3t\right)^{2}.

$$\frac{180t+124+900-180t+9t^{2}}{25}=\left(10-2t\right)^{2}$$

Combine like terms in 180t+124+900-180t+9t^{2}.

$$\frac{1024+9t^{2}}{25}=\left(10-2t\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(10-2t\right)^{2}.

$$\frac{1024+9t^{2}}{25}=100-40t+4t^{2}$$

Divide each term of 1024+9t^{2} by 25 to get \frac{1024}{25}+\frac{9}{25}t^{2}.

$$\frac{1024}{25}+\frac{9}{25}t^{2}=100-40t+4t^{2}$$

Subtract 100 from both sides.

$$\frac{1024}{25}+\frac{9}{25}t^{2}-100=-40t+4t^{2}$$

Subtract 100 from \frac{1024}{25} to get -\frac{1476}{25}.

$$-\frac{1476}{25}+\frac{9}{25}t^{2}=-40t+4t^{2}$$

Add 40t to both sides.

$$-\frac{1476}{25}+\frac{9}{25}t^{2}+40t=4t^{2}$$

Subtract 4t^{2} from both sides.

$$-\frac{1476}{25}+\frac{9}{25}t^{2}+40t-4t^{2}=0$$

Combine \frac{9}{25}t^{2} and -4t^{2} to get -\frac{91}{25}t^{2}.

$$-\frac{1476}{25}-\frac{91}{25}t^{2}+40t=0$$

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.

$$-\frac{91}{25}t^{2}+40t-\frac{1476}{25}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{91}{25} for a, 40 for b, and -\frac{1476}{25} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-40±\sqrt{40^{2}-4\left(-\frac{91}{25}\right)\left(-\frac{1476}{25}\right)}}{2\left(-\frac{91}{25}\right)}$$

Square 40.

$$t=\frac{-40±\sqrt{1600-4\left(-\frac{91}{25}\right)\left(-\frac{1476}{25}\right)}}{2\left(-\frac{91}{25}\right)}$$

Multiply -4 times -\frac{91}{25}.

$$t=\frac{-40±\sqrt{1600+\frac{364}{25}\left(-\frac{1476}{25}\right)}}{2\left(-\frac{91}{25}\right)}$$

Multiply \frac{364}{25} times -\frac{1476}{25} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$t=\frac{-40±\sqrt{1600-\frac{537264}{625}}}{2\left(-\frac{91}{25}\right)}$$

Add 1600 to -\frac{537264}{625}.

$$t=\frac{-40±\sqrt{\frac{462736}{625}}}{2\left(-\frac{91}{25}\right)}$$

Take the square root of \frac{462736}{625}.

$$t=\frac{-40±\frac{4\sqrt{28921}}{25}}{2\left(-\frac{91}{25}\right)}$$

Multiply 2 times -\frac{91}{25}.

$$t=\frac{-40±\frac{4\sqrt{28921}}{25}}{-\frac{182}{25}}$$

Now solve the equation t=\frac{-40±\frac{4\sqrt{28921}}{25}}{-\frac{182}{25}} when ± is plus. Add -40 to \frac{4\sqrt{28921}}{25}.

$$t=\frac{\frac{4\sqrt{28921}}{25}-40}{-\frac{182}{25}}$$

Divide -40+\frac{4\sqrt{28921}}{25} by -\frac{182}{25} by multiplying -40+\frac{4\sqrt{28921}}{25} by the reciprocal of -\frac{182}{25}.

$$t=\frac{500-2\sqrt{28921}}{91}$$

Now solve the equation t=\frac{-40±\frac{4\sqrt{28921}}{25}}{-\frac{182}{25}} when ± is minus. Subtract \frac{4\sqrt{28921}}{25} from -40.

$$t=\frac{-\frac{4\sqrt{28921}}{25}-40}{-\frac{182}{25}}$$

Divide -40-\frac{4\sqrt{28921}}{25} by -\frac{182}{25} by multiplying -40-\frac{4\sqrt{28921}}{25} by the reciprocal of -\frac{182}{25}.

$$t=\frac{2\sqrt{28921}+500}{91}$$

The equation is now solved.

$$t=\frac{500-2\sqrt{28921}}{91} t=\frac{2\sqrt{28921}+500}{91}$$

Steps for Completing the Square

Subtract 2 from 10 to get 8.

$$\left(\frac{4\times 8}{5}\right)^{2}+\left(6-\frac{3\left(10-t\right)}{5}\right)^{2}=\left(10-2t\right)^{2}$$

Multiply 4 and 8 to get 32.

$$\left(\frac{32}{5}\right)^{2}+\left(6-\frac{3\left(10-t\right)}{5}\right)^{2}=\left(10-2t\right)^{2}$$

Calculate \frac{32}{5} to the power of 2 and get \frac{1024}{25}.

$$\frac{1024}{25}+\left(6-\frac{3\left(10-t\right)}{5}\right)^{2}=\left(10-2t\right)^{2}$$

Use the distributive property to multiply 3 by 10-t.

$$\frac{1024}{25}+\left(6-\frac{30-3t}{5}\right)^{2}=\left(10-2t\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(6-\frac{30-3t}{5}\right)^{2}.

$$\frac{1024}{25}+36+12\left(-\frac{30-3t}{5}\right)+\left(-\frac{30-3t}{5}\right)^{2}=\left(10-2t\right)^{2}$$

Express 12\left(-\frac{30-3t}{5}\right) as a single fraction.

$$\frac{1024}{25}+36+\frac{-12\left(30-3t\right)}{5}+\left(-\frac{30-3t}{5}\right)^{2}=\left(10-2t\right)^{2}$$

Calculate -\frac{30-3t}{5} to the power of 2 and get \left(\frac{30-3t}{5}\right)^{2}.

$$\frac{1024}{25}+36+\frac{-12\left(30-3t\right)}{5}+\left(\frac{30-3t}{5}\right)^{2}=\left(10-2t\right)^{2}$$

Add \frac{1024}{25} and 36 to get \frac{1924}{25}.

$$\frac{1924}{25}+\frac{-12\left(30-3t\right)}{5}+\left(\frac{30-3t}{5}\right)^{2}=\left(10-2t\right)^{2}$$

To raise \frac{30-3t}{5} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{1924}{25}+\frac{-12\left(30-3t\right)}{5}+\frac{\left(30-3t\right)^{2}}{5^{2}}=\left(10-2t\right)^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 25 and 5 is 25. Multiply \frac{-12\left(30-3t\right)}{5} times \frac{5}{5}.

$$\frac{1924}{25}+\frac{5\left(-12\right)\left(30-3t\right)}{25}+\frac{\left(30-3t\right)^{2}}{5^{2}}=\left(10-2t\right)^{2}$$

Since \frac{1924}{25} and \frac{5\left(-12\right)\left(30-3t\right)}{25} have the same denominator, add them by adding their numerators.

$$\frac{1924+5\left(-12\right)\left(30-3t\right)}{25}+\frac{\left(30-3t\right)^{2}}{5^{2}}=\left(10-2t\right)^{2}$$

Do the multiplications in 1924+5\left(-12\right)\left(30-3t\right).

$$\frac{1924-1800+180t}{25}+\frac{\left(30-3t\right)^{2}}{5^{2}}=\left(10-2t\right)^{2}$$

Combine like terms in 1924-1800+180t.

$$\frac{180t+124}{25}+\frac{\left(30-3t\right)^{2}}{5^{2}}=\left(10-2t\right)^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Expand 5^{2}.

$$\frac{180t+124}{25}+\frac{\left(30-3t\right)^{2}}{25}=\left(10-2t\right)^{2}$$

Since \frac{180t+124}{25} and \frac{\left(30-3t\right)^{2}}{25} have the same denominator, add them by adding their numerators.

$$\frac{180t+124+\left(30-3t\right)^{2}}{25}=\left(10-2t\right)^{2}$$

Do the multiplications in 180t+124+\left(30-3t\right)^{2}.

$$\frac{180t+124+900-180t+9t^{2}}{25}=\left(10-2t\right)^{2}$$

Combine like terms in 180t+124+900-180t+9t^{2}.

$$\frac{1024+9t^{2}}{25}=\left(10-2t\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(10-2t\right)^{2}.

$$\frac{1024+9t^{2}}{25}=100-40t+4t^{2}$$

Divide each term of 1024+9t^{2} by 25 to get \frac{1024}{25}+\frac{9}{25}t^{2}.

$$\frac{1024}{25}+\frac{9}{25}t^{2}=100-40t+4t^{2}$$

Add 40t to both sides.

$$\frac{1024}{25}+\frac{9}{25}t^{2}+40t=100+4t^{2}$$

Subtract 4t^{2} from both sides.

$$\frac{1024}{25}+\frac{9}{25}t^{2}+40t-4t^{2}=100$$

Combine \frac{9}{25}t^{2} and -4t^{2} to get -\frac{91}{25}t^{2}.

$$\frac{1024}{25}-\frac{91}{25}t^{2}+40t=100$$

Subtract \frac{1024}{25} from both sides.

$$-\frac{91}{25}t^{2}+40t=100-\frac{1024}{25}$$

Subtract \frac{1024}{25} from 100 to get \frac{1476}{25}.

$$-\frac{91}{25}t^{2}+40t=\frac{1476}{25}$$

Divide both sides of the equation by -\frac{91}{25}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{91}{25}t^{2}+40t}{-\frac{91}{25}}=\frac{\frac{1476}{25}}{-\frac{91}{25}}$$

Dividing by -\frac{91}{25} undoes the multiplication by -\frac{91}{25}.

$$t^{2}+\frac{40}{-\frac{91}{25}}t=\frac{\frac{1476}{25}}{-\frac{91}{25}}$$

Divide 40 by -\frac{91}{25} by multiplying 40 by the reciprocal of -\frac{91}{25}.

$$t^{2}-\frac{1000}{91}t=\frac{\frac{1476}{25}}{-\frac{91}{25}}$$

Divide \frac{1476}{25} by -\frac{91}{25} by multiplying \frac{1476}{25} by the reciprocal of -\frac{91}{25}.

$$t^{2}-\frac{1000}{91}t=-\frac{1476}{91}$$

Divide -\frac{1000}{91}, the coefficient of the x term, by 2 to get -\frac{500}{91}. Then add the square of -\frac{500}{91} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-\frac{1000}{91}t+\left(-\frac{500}{91}\right)^{2}=-\frac{1476}{91}+\left(-\frac{500}{91}\right)^{2}$$

Square -\frac{500}{91} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{1000}{91}t+\frac{250000}{8281}=-\frac{1476}{91}+\frac{250000}{8281}$$

Add -\frac{1476}{91} to \frac{250000}{8281} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}-\frac{1000}{91}t+\frac{250000}{8281}=\frac{115684}{8281}$$

Factor t^{2}-\frac{1000}{91}t+\frac{250000}{8281}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-\frac{500}{91}\right)^{2}=\frac{115684}{8281}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{500}{91}\right)^{2}}=\sqrt{\frac{115684}{8281}}$$

Simplify.

$$t-\frac{500}{91}=\frac{2\sqrt{28921}}{91} t-\frac{500}{91}=-\frac{2\sqrt{28921}}{91}$$

Add \frac{500}{91} to both sides of the equation.

$$t=\frac{2\sqrt{28921}+500}{91} t=\frac{500-2\sqrt{28921}}{91}$$