Solve (1)(12-2t)^2+t^2=68

$$( 1 ) ( 12 - 2 t ) ^ { 2 } + t ^ { 2 } = 68$$

Solve for t

$$t=2$$

$$t = \frac{38}{5} = 7\frac{3}{5} = 7.6$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(12-2t\right)^{2}.

$$1\left(144-48t+4t^{2}\right)+t^{2}=68$$

Use the distributive property to multiply 1 by 144-48t+4t^{2}.

$$144-48t+4t^{2}+t^{2}=68$$

Combine 4t^{2} and t^{2} to get 5t^{2}.

$$144-48t+5t^{2}=68$$

Subtract 68 from both sides.

$$144-48t+5t^{2}-68=0$$

Subtract 68 from 144 to get 76.

$$76-48t+5t^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$5t^{2}-48t+76=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5t^{2}+at+bt+76. To find a and b, set up a system to be solved.

$$a+b=-48 ab=5\times 76=380$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 380.

$$-1,-380 -2,-190 -4,-95 -5,-76 -10,-38 -19,-20$$

Calculate the sum for each pair.

$$-1-380=-381 -2-190=-192 -4-95=-99 -5-76=-81 -10-38=-48 -19-20=-39$$

The solution is the pair that gives sum -48.

$$a=-38 b=-10$$

Rewrite 5t^{2}-48t+76 as \left(5t^{2}-38t\right)+\left(-10t+76\right).

$$\left(5t^{2}-38t\right)+\left(-10t+76\right)$$

Factor out t in the first and -2 in the second group.

$$t\left(5t-38\right)-2\left(5t-38\right)$$

Factor out common term 5t-38 by using distributive property.

$$\left(5t-38\right)\left(t-2\right)$$

To find equation solutions, solve 5t-38=0 and t-2=0.

$$t=\frac{38}{5} t=2$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(12-2t\right)^{2}.

$$1\left(144-48t+4t^{2}\right)+t^{2}=68$$

Use the distributive property to multiply 1 by 144-48t+4t^{2}.

$$144-48t+4t^{2}+t^{2}=68$$

Combine 4t^{2} and t^{2} to get 5t^{2}.

$$144-48t+5t^{2}=68$$

Subtract 68 from both sides.

$$144-48t+5t^{2}-68=0$$

Subtract 68 from 144 to get 76.

$$76-48t+5t^{2}=0$$

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.

$$5t^{2}-48t+76=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -48 for b, and 76 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-48\right)±\sqrt{\left(-48\right)^{2}-4\times 5\times 76}}{2\times 5}$$

Square -48.

$$t=\frac{-\left(-48\right)±\sqrt{2304-4\times 5\times 76}}{2\times 5}$$

Multiply -4 times 5.

$$t=\frac{-\left(-48\right)±\sqrt{2304-20\times 76}}{2\times 5}$$

Multiply -20 times 76.

$$t=\frac{-\left(-48\right)±\sqrt{2304-1520}}{2\times 5}$$

Add 2304 to -1520.

$$t=\frac{-\left(-48\right)±\sqrt{784}}{2\times 5}$$

Take the square root of 784.

$$t=\frac{-\left(-48\right)±28}{2\times 5}$$

The opposite of -48 is 48.

$$t=\frac{48±28}{2\times 5}$$

Multiply 2 times 5.

$$t=\frac{48±28}{10}$$

Now solve the equation t=\frac{48±28}{10} when ± is plus. Add 48 to 28.

$$t=\frac{76}{10}$$

Reduce the fraction \frac{76}{10} to lowest terms by extracting and canceling out 2.

$$t=\frac{38}{5}$$

Now solve the equation t=\frac{48±28}{10} when ± is minus. Subtract 28 from 48.

$$t=\frac{20}{10}$$

Divide 20 by 10.

$$t=2$$

The equation is now solved.

$$t=\frac{38}{5} t=2$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(12-2t\right)^{2}.

$$1\left(144-48t+4t^{2}\right)+t^{2}=68$$

Use the distributive property to multiply 1 by 144-48t+4t^{2}.

$$144-48t+4t^{2}+t^{2}=68$$

Combine 4t^{2} and t^{2} to get 5t^{2}.

$$144-48t+5t^{2}=68$$

Subtract 144 from both sides.

$$-48t+5t^{2}=68-144$$

Subtract 144 from 68 to get -76.

$$-48t+5t^{2}=-76$$

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.

$$5t^{2}-48t=-76$$

Divide both sides by 5.

$$\frac{5t^{2}-48t}{5}=-\frac{76}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$t^{2}-\frac{48}{5}t=-\frac{76}{5}$$

Divide -\frac{48}{5}, the coefficient of the x term, by 2 to get -\frac{24}{5}. Then add the square of -\frac{24}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-\frac{48}{5}t+\left(-\frac{24}{5}\right)^{2}=-\frac{76}{5}+\left(-\frac{24}{5}\right)^{2}$$

Square -\frac{24}{5} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{48}{5}t+\frac{576}{25}=-\frac{76}{5}+\frac{576}{25}$$

Add -\frac{76}{5} to \frac{576}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}-\frac{48}{5}t+\frac{576}{25}=\frac{196}{25}$$

Factor t^{2}-\frac{48}{5}t+\frac{576}{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}.

$$\left(t-\frac{24}{5}\right)^{2}=\frac{196}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{24}{5}\right)^{2}}=\sqrt{\frac{196}{25}}$$

Simplify.

$$t-\frac{24}{5}=\frac{14}{5} t-\frac{24}{5}=-\frac{14}{5}$$

Add \frac{24}{5} to both sides of the equation.

$$t=\frac{38}{5} t=2$$