Solve 3=48-1/2*8*(2t-8)-1/2*6*(t-6)-1/2*(14-t)*(14-2t)

$$3 = 48 - \frac { 1 } { 2 } \times 8 \times ( 2 t - 8 ) - \frac { 1 } { 2 } \times 6 \times ( t - 6 ) - \frac { 1 } { 2 } \times ( 14 - t ) \times ( 14 - 2 t )$$

Solve for t

$$t=\sqrt{22}+5\approx 9.69041576$$

$$t=5-\sqrt{22}\approx 0.30958424$$

Steps Using the Quadratic Formula

Multiply \frac{1}{2} and 8 to get 4.

$$3=48-4\left(2t-8\right)-\frac{1}{2}\times 6\left(t-6\right)-\frac{1}{2}\left(14-t\right)\left(14-2t\right)$$

Use the distributive property to multiply -4 by 2t-8.

$$3=48-8t+32-\frac{1}{2}\times 6\left(t-6\right)-\frac{1}{2}\left(14-t\right)\left(14-2t\right)$$

Add 48 and 32 to get 80.

$$3=80-8t-\frac{1}{2}\times 6\left(t-6\right)-\frac{1}{2}\left(14-t\right)\left(14-2t\right)$$

Multiply \frac{1}{2} and 6 to get 3.

$$3=80-8t-3\left(t-6\right)-\frac{1}{2}\left(14-t\right)\left(14-2t\right)$$

Use the distributive property to multiply -3 by t-6.

$$3=80-8t-3t+18-\frac{1}{2}\left(14-t\right)\left(14-2t\right)$$

Combine -8t and -3t to get -11t.

$$3=80-11t+18-\frac{1}{2}\left(14-t\right)\left(14-2t\right)$$

Add 80 and 18 to get 98.

$$3=98-11t-\frac{1}{2}\left(14-t\right)\left(14-2t\right)$$

Swap sides so that all variable terms are on the left hand side.

$$98-11t-\frac{1}{2}\left(14-t\right)\left(14-2t\right)=3$$

Subtract 3 from both sides.

$$98-11t-\frac{1}{2}\left(14-t\right)\left(14-2t\right)-3=0$$

Use the distributive property to multiply -\frac{1}{2} by 14-t.

$$98-11t+\left(-7+\frac{1}{2}t\right)\left(14-2t\right)-3=0$$

Use the distributive property to multiply -7+\frac{1}{2}t by 14-2t and combine like terms.

$$98-11t-98+21t-t^{2}-3=0$$

Subtract 98 from 98 to get 0.

$$-11t+21t-t^{2}-3=0$$

Combine -11t and 21t to get 10t.

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 10 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-10±\sqrt{10^{2}-4\left(-1\right)\left(-3\right)}}{2\left(-1\right)}$$

Square 10.

$$t=\frac{-10±\sqrt{100-4\left(-1\right)\left(-3\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$t=\frac{-10±\sqrt{100+4\left(-3\right)}}{2\left(-1\right)}$$

Multiply 4 times -3.

$$t=\frac{-10±\sqrt{100-12}}{2\left(-1\right)}$$

Add 100 to -12.

$$t=\frac{-10±\sqrt{88}}{2\left(-1\right)}$$

Take the square root of 88.

$$t=\frac{-10±2\sqrt{22}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$t=\frac{-10±2\sqrt{22}}{-2}$$

Now solve the equation t=\frac{-10±2\sqrt{22}}{-2} when ± is plus. Add -10 to 2\sqrt{22}.

$$t=\frac{2\sqrt{22}-10}{-2}$$

Divide -10+2\sqrt{22} by -2.

$$t=5-\sqrt{22}$$

Now solve the equation t=\frac{-10±2\sqrt{22}}{-2} when ± is minus. Subtract 2\sqrt{22} from -10.

$$t=\frac{-2\sqrt{22}-10}{-2}$$

Divide -10-2\sqrt{22} by -2.

$$t=\sqrt{22}+5$$

The equation is now solved.

$$t=5-\sqrt{22} t=\sqrt{22}+5$$

Steps for Completing the Square

Multiply \frac{1}{2} and 8 to get 4.

$$3=48-4\left(2t-8\right)-\frac{1}{2}\times 6\left(t-6\right)-\frac{1}{2}\left(14-t\right)\left(14-2t\right)$$

Use the distributive property to multiply -4 by 2t-8.

$$3=48-8t+32-\frac{1}{2}\times 6\left(t-6\right)-\frac{1}{2}\left(14-t\right)\left(14-2t\right)$$

Add 48 and 32 to get 80.

$$3=80-8t-\frac{1}{2}\times 6\left(t-6\right)-\frac{1}{2}\left(14-t\right)\left(14-2t\right)$$

Multiply \frac{1}{2} and 6 to get 3.

$$3=80-8t-3\left(t-6\right)-\frac{1}{2}\left(14-t\right)\left(14-2t\right)$$

Use the distributive property to multiply -3 by t-6.

$$3=80-8t-3t+18-\frac{1}{2}\left(14-t\right)\left(14-2t\right)$$

Combine -8t and -3t to get -11t.

$$3=80-11t+18-\frac{1}{2}\left(14-t\right)\left(14-2t\right)$$

Add 80 and 18 to get 98.

$$3=98-11t-\frac{1}{2}\left(14-t\right)\left(14-2t\right)$$

Swap sides so that all variable terms are on the left hand side.

$$98-11t-\frac{1}{2}\left(14-t\right)\left(14-2t\right)=3$$

Use the distributive property to multiply -\frac{1}{2} by 14-t.

$$98-11t+\left(-7+\frac{1}{2}t\right)\left(14-2t\right)=3$$

Use the distributive property to multiply -7+\frac{1}{2}t by 14-2t and combine like terms.

$$98-11t-98+21t-t^{2}=3$$

Subtract 98 from 98 to get 0.

$$-11t+21t-t^{2}=3$$

Combine -11t and 21t to get 10t.

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

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.

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

Divide both sides by -1.

$$\frac{-t^{2}+10t}{-1}=\frac{3}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$t^{2}+\frac{10}{-1}t=\frac{3}{-1}$$

Divide 10 by -1.

$$t^{2}-10t=\frac{3}{-1}$$

Divide 3 by -1.

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

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -5.

$$t^{2}-10t+25=-3+25$$

Add -3 to 25.

$$t^{2}-10t+25=22$$

Factor t^{2}-10t+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}.

$$\left(t-5\right)^{2}=22$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-5\right)^{2}}=\sqrt{22}$$

Simplify.

$$t-5=\sqrt{22} t-5=-\sqrt{22}$$

Add 5 to both sides of the equation.

$$t=\sqrt{22}+5 t=5-\sqrt{22}$$