Solve 24-[4(6-2t)-2t(4-t)-6t]21/2=16

$$24 - [ 4 ( 6 - 2 t ) - 2 t ( 4 - t ) - 6 t ] 2 \frac { 1 } { 2 } = 16$$

Solve for t

$$t=\frac{\sqrt{1985}}{10}+\frac{11}{2}\approx 9.955333882$$

$$t=-\frac{\sqrt{1985}}{10}+\frac{11}{2}\approx 1.044666118$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

$$48-\left(4\left(6-2t\right)-2t\left(4-t\right)-6t\right)\left(2\times 2+1\right)=32$$

Use the distributive property to multiply 4 by 6-2t.

$$48-\left(24-8t-2t\left(4-t\right)-6t\right)\left(2\times 2+1\right)=32$$

Multiply 2 and 2 to get 4.

$$48-\left(24-8t-2t\left(4-t\right)-6t\right)\left(4+1\right)=32$$

Add 4 and 1 to get 5.

$$48-\left(24-8t-2t\left(4-t\right)-6t\right)\times 5=32$$

Use the distributive property to multiply 24-8t-2t\left(4-t\right)-6t by 5.

$$48-\left(5\left(24-8t-2t\left(4-t\right)\right)-30t\right)=32$$

To find the opposite of 5\left(24-8t-2t\left(4-t\right)\right)-30t, find the opposite of each term.

$$48-5\left(24-8t-2t\left(4-t\right)\right)-\left(-30t\right)=32$$

The opposite of -30t is 30t.

$$48-5\left(24-8t-2t\left(4-t\right)\right)+30t=32$$

Subtract 32 from both sides.

$$48-5\left(24-8t-2t\left(4-t\right)\right)+30t-32=0$$

Subtract 32 from 48 to get 16.

$$16-5\left(24-8t-2t\left(4-t\right)\right)+30t=0$$

Subtract 16 from both sides. Anything subtracted from zero gives its negation.

$$-5\left(24-8t-2t\left(4-t\right)\right)+30t=-16$$

Add 16 to both sides.

$$-5\left(24-8t-2t\left(4-t\right)\right)+30t+16=0$$

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

$$-5\left(24-8t-8t+2t^{2}\right)+30t+16=0$$

Combine -8t and -8t to get -16t.

$$-5\left(24-16t+2t^{2}\right)+30t+16=0$$

Use the distributive property to multiply -5 by 24-16t+2t^{2}.

$$-120+80t-10t^{2}+30t+16=0$$

Combine 80t and 30t to get 110t.

$$-120+110t-10t^{2}+16=0$$

Add -120 and 16 to get -104.

$$-104+110t-10t^{2}=0$$

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.

$$-10t^{2}+110t-104=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -10 for a, 110 for b, and -104 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-110±\sqrt{110^{2}-4\left(-10\right)\left(-104\right)}}{2\left(-10\right)}$$

Square 110.

$$t=\frac{-110±\sqrt{12100-4\left(-10\right)\left(-104\right)}}{2\left(-10\right)}$$

Multiply -4 times -10.

$$t=\frac{-110±\sqrt{12100+40\left(-104\right)}}{2\left(-10\right)}$$

Multiply 40 times -104.

$$t=\frac{-110±\sqrt{12100-4160}}{2\left(-10\right)}$$

Add 12100 to -4160.

$$t=\frac{-110±\sqrt{7940}}{2\left(-10\right)}$$

Take the square root of 7940.

$$t=\frac{-110±2\sqrt{1985}}{2\left(-10\right)}$$

Multiply 2 times -10.

$$t=\frac{-110±2\sqrt{1985}}{-20}$$

Now solve the equation t=\frac{-110±2\sqrt{1985}}{-20} when ± is plus. Add -110 to 2\sqrt{1985}.

$$t=\frac{2\sqrt{1985}-110}{-20}$$

Divide -110+2\sqrt{1985} by -20.

$$t=-\frac{\sqrt{1985}}{10}+\frac{11}{2}$$

Now solve the equation t=\frac{-110±2\sqrt{1985}}{-20} when ± is minus. Subtract 2\sqrt{1985} from -110.

$$t=\frac{-2\sqrt{1985}-110}{-20}$$

Divide -110-2\sqrt{1985} by -20.

$$t=\frac{\sqrt{1985}}{10}+\frac{11}{2}$$

The equation is now solved.

$$t=-\frac{\sqrt{1985}}{10}+\frac{11}{2} t=\frac{\sqrt{1985}}{10}+\frac{11}{2}$$

Steps for Completing the Square

Multiply both sides of the equation by 2.

$$48-\left(4\left(6-2t\right)-2t\left(4-t\right)-6t\right)\left(2\times 2+1\right)=32$$

Use the distributive property to multiply 4 by 6-2t.

$$48-\left(24-8t-2t\left(4-t\right)-6t\right)\left(2\times 2+1\right)=32$$

Multiply 2 and 2 to get 4.

$$48-\left(24-8t-2t\left(4-t\right)-6t\right)\left(4+1\right)=32$$

Add 4 and 1 to get 5.

$$48-\left(24-8t-2t\left(4-t\right)-6t\right)\times 5=32$$

Use the distributive property to multiply 24-8t-2t\left(4-t\right)-6t by 5.

$$48-\left(5\left(24-8t-2t\left(4-t\right)\right)-30t\right)=32$$

To find the opposite of 5\left(24-8t-2t\left(4-t\right)\right)-30t, find the opposite of each term.

$$48-5\left(24-8t-2t\left(4-t\right)\right)-\left(-30t\right)=32$$

The opposite of -30t is 30t.

$$48-5\left(24-8t-2t\left(4-t\right)\right)+30t=32$$

Subtract 48 from both sides.

$$-5\left(24-8t-2t\left(4-t\right)\right)+30t=32-48$$

Subtract 48 from 32 to get -16.

$$-5\left(24-8t-2t\left(4-t\right)\right)+30t=-16$$

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

$$-5\left(24-8t-8t+2t^{2}\right)+30t=-16$$

Combine -8t and -8t to get -16t.

$$-5\left(24-16t+2t^{2}\right)+30t=-16$$

Use the distributive property to multiply -5 by 24-16t+2t^{2}.

$$-120+80t-10t^{2}+30t=-16$$

Combine 80t and 30t to get 110t.

$$-120+110t-10t^{2}=-16$$

Add 120 to both sides.

$$110t-10t^{2}=-16+120$$

Add -16 and 120 to get 104.

$$110t-10t^{2}=104$$

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.

$$-10t^{2}+110t=104$$

Divide both sides by -10.

$$\frac{-10t^{2}+110t}{-10}=\frac{104}{-10}$$

Dividing by -10 undoes the multiplication by -10.

$$t^{2}+\frac{110}{-10}t=\frac{104}{-10}$$

Divide 110 by -10.

$$t^{2}-11t=\frac{104}{-10}$$

Reduce the fraction \frac{104}{-10} to lowest terms by extracting and canceling out 2.

$$t^{2}-11t=-\frac{52}{5}$$

Divide -11, the coefficient of the x term, by 2 to get -\frac{11}{2}. Then add the square of -\frac{11}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-11t+\left(-\frac{11}{2}\right)^{2}=-\frac{52}{5}+\left(-\frac{11}{2}\right)^{2}$$

Square -\frac{11}{2} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-11t+\frac{121}{4}=-\frac{52}{5}+\frac{121}{4}$$

Add -\frac{52}{5} to \frac{121}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}-11t+\frac{121}{4}=\frac{397}{20}$$

Factor t^{2}-11t+\frac{121}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-\frac{11}{2}\right)^{2}=\frac{397}{20}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{11}{2}\right)^{2}}=\sqrt{\frac{397}{20}}$$

Simplify.

$$t-\frac{11}{2}=\frac{\sqrt{1985}}{10} t-\frac{11}{2}=-\frac{\sqrt{1985}}{10}$$

Add \frac{11}{2} to both sides of the equation.

$$t=\frac{\sqrt{1985}}{10}+\frac{11}{2} t=-\frac{\sqrt{1985}}{10}+\frac{11}{2}$$