Solve 200=50t+6/2t^2

$$200 = 50 t + \frac { 6 } { 2 } t ^ { 2 }$$

Solve for t

$$t = \frac{10}{3} = 3\frac{1}{3} \approx 3.333333333$$

$$t=-20$$

Steps Using Factoring By Grouping

Divide 6 by 2 to get 3.

$$200=50t+3t^{2}$$

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

$$50t+3t^{2}=200$$

Subtract 200 from both sides.

$$50t+3t^{2}-200=0$$

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

$$3t^{2}+50t-200=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3t^{2}+at+bt-200. To find a and b, set up a system to be solved.

$$a+b=50 ab=3\left(-200\right)=-600$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -600.

$$-1,600 -2,300 -3,200 -4,150 -5,120 -6,100 -8,75 -10,60 -12,50 -15,40 -20,30 -24,25$$

Calculate the sum for each pair.

$$-1+600=599 -2+300=298 -3+200=197 -4+150=146 -5+120=115 -6+100=94 -8+75=67 -10+60=50 -12+50=38 -15+40=25 -20+30=10 -24+25=1$$

The solution is the pair that gives sum 50.

$$a=-10 b=60$$

Rewrite 3t^{2}+50t-200 as \left(3t^{2}-10t\right)+\left(60t-200\right).

$$\left(3t^{2}-10t\right)+\left(60t-200\right)$$

Factor out t in the first and 20 in the second group.

$$t\left(3t-10\right)+20\left(3t-10\right)$$

Factor out common term 3t-10 by using distributive property.

$$\left(3t-10\right)\left(t+20\right)$$

To find equation solutions, solve 3t-10=0 and t+20=0.

$$t=\frac{10}{3} t=-20$$

Steps Using the Quadratic Formula

Divide 6 by 2 to get 3.

$$200=50t+3t^{2}$$

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

$$50t+3t^{2}=200$$

Subtract 200 from both sides.

$$50t+3t^{2}-200=0$$

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.

$$3t^{2}+50t-200=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 50 for b, and -200 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-50±\sqrt{50^{2}-4\times 3\left(-200\right)}}{2\times 3}$$

Square 50.

$$t=\frac{-50±\sqrt{2500-4\times 3\left(-200\right)}}{2\times 3}$$

Multiply -4 times 3.

$$t=\frac{-50±\sqrt{2500-12\left(-200\right)}}{2\times 3}$$

Multiply -12 times -200.

$$t=\frac{-50±\sqrt{2500+2400}}{2\times 3}$$

Add 2500 to 2400.

$$t=\frac{-50±\sqrt{4900}}{2\times 3}$$

Take the square root of 4900.

$$t=\frac{-50±70}{2\times 3}$$

Multiply 2 times 3.

$$t=\frac{-50±70}{6}$$

Now solve the equation t=\frac{-50±70}{6} when ± is plus. Add -50 to 70.

$$t=\frac{20}{6}$$

Reduce the fraction \frac{20}{6} to lowest terms by extracting and canceling out 2.

$$t=\frac{10}{3}$$

Now solve the equation t=\frac{-50±70}{6} when ± is minus. Subtract 70 from -50.

$$t=-\frac{120}{6}$$

Divide -120 by 6.

$$t=-20$$

The equation is now solved.

$$t=\frac{10}{3} t=-20$$

Steps for Completing the Square

Divide 6 by 2 to get 3.

$$200=50t+3t^{2}$$

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

$$50t+3t^{2}=200$$

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.

$$3t^{2}+50t=200$$

Divide both sides by 3.

$$\frac{3t^{2}+50t}{3}=\frac{200}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$t^{2}+\frac{50}{3}t=\frac{200}{3}$$

Divide \frac{50}{3}, the coefficient of the x term, by 2 to get \frac{25}{3}. Then add the square of \frac{25}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}+\frac{50}{3}t+\left(\frac{25}{3}\right)^{2}=\frac{200}{3}+\left(\frac{25}{3}\right)^{2}$$

Square \frac{25}{3} by squaring both the numerator and the denominator of the fraction.

$$t^{2}+\frac{50}{3}t+\frac{625}{9}=\frac{200}{3}+\frac{625}{9}$$

Add \frac{200}{3} to \frac{625}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}+\frac{50}{3}t+\frac{625}{9}=\frac{1225}{9}$$

Factor t^{2}+\frac{50}{3}t+\frac{625}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t+\frac{25}{3}\right)^{2}=\frac{1225}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t+\frac{25}{3}\right)^{2}}=\sqrt{\frac{1225}{9}}$$

Simplify.

$$t+\frac{25}{3}=\frac{35}{3} t+\frac{25}{3}=-\frac{35}{3}$$

Subtract \frac{25}{3} from both sides of the equation.

$$t=\frac{10}{3} t=-20$$