Solve 30-11t+t^2/6=10t-2t^2/15

$$\frac { 30 - 11 t + t ^ { 2 } } { 6 } = \frac { 10 t - 2 t ^ { 2 } } { 15 }$$

Solve for t

$$t=5$$

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

Steps Using the Quadratic Formula

Multiply both sides of the equation by 30, the least common multiple of 6,15.

$$5\left(30-11t+t^{2}\right)=2\left(10t-2t^{2}\right)$$

Use the distributive property to multiply 5 by 30-11t+t^{2}.

$$150-55t+5t^{2}=2\left(10t-2t^{2}\right)$$

Use the distributive property to multiply 2 by 10t-2t^{2}.

$$150-55t+5t^{2}=20t-4t^{2}$$

Subtract 20t from both sides.

$$150-55t+5t^{2}-20t=-4t^{2}$$

Combine -55t and -20t to get -75t.

$$150-75t+5t^{2}=-4t^{2}$$

Add 4t^{2} to both sides.

$$150-75t+5t^{2}+4t^{2}=0$$

Combine 5t^{2} and 4t^{2} to get 9t^{2}.

$$150-75t+9t^{2}=0$$

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.

$$9t^{2}-75t+150=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -75 for b, and 150 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-75\right)±\sqrt{\left(-75\right)^{2}-4\times 9\times 150}}{2\times 9}$$

Square -75.

$$t=\frac{-\left(-75\right)±\sqrt{5625-4\times 9\times 150}}{2\times 9}$$

Multiply -4 times 9.

$$t=\frac{-\left(-75\right)±\sqrt{5625-36\times 150}}{2\times 9}$$

Multiply -36 times 150.

$$t=\frac{-\left(-75\right)±\sqrt{5625-5400}}{2\times 9}$$

Add 5625 to -5400.

$$t=\frac{-\left(-75\right)±\sqrt{225}}{2\times 9}$$

Take the square root of 225.

$$t=\frac{-\left(-75\right)±15}{2\times 9}$$

The opposite of -75 is 75.

$$t=\frac{75±15}{2\times 9}$$

Multiply 2 times 9.

$$t=\frac{75±15}{18}$$

Now solve the equation t=\frac{75±15}{18} when ± is plus. Add 75 to 15.

$$t=\frac{90}{18}$$

Divide 90 by 18.

$$t=5$$

Now solve the equation t=\frac{75±15}{18} when ± is minus. Subtract 15 from 75.

$$t=\frac{60}{18}$$

Reduce the fraction \frac{60}{18} to lowest terms by extracting and canceling out 6.

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

The equation is now solved.

$$t=5 t=\frac{10}{3}$$

Steps for Completing the Square

Multiply both sides of the equation by 30, the least common multiple of 6,15.

$$5\left(30-11t+t^{2}\right)=2\left(10t-2t^{2}\right)$$

Use the distributive property to multiply 5 by 30-11t+t^{2}.

$$150-55t+5t^{2}=2\left(10t-2t^{2}\right)$$

Use the distributive property to multiply 2 by 10t-2t^{2}.

$$150-55t+5t^{2}=20t-4t^{2}$$

Subtract 20t from both sides.

$$150-55t+5t^{2}-20t=-4t^{2}$$

Combine -55t and -20t to get -75t.

$$150-75t+5t^{2}=-4t^{2}$$

Add 4t^{2} to both sides.

$$150-75t+5t^{2}+4t^{2}=0$$

Combine 5t^{2} and 4t^{2} to get 9t^{2}.

$$150-75t+9t^{2}=0$$

Subtract 150 from both sides. Anything subtracted from zero gives its negation.

$$-75t+9t^{2}=-150$$

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.

$$9t^{2}-75t=-150$$

Divide both sides by 9.

$$\frac{9t^{2}-75t}{9}=-\frac{150}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$t^{2}+\left(-\frac{75}{9}\right)t=-\frac{150}{9}$$

Reduce the fraction \frac{-75}{9} to lowest terms by extracting and canceling out 3.

$$t^{2}-\frac{25}{3}t=-\frac{150}{9}$$

Reduce the fraction \frac{-150}{9} to lowest terms by extracting and canceling out 3.

$$t^{2}-\frac{25}{3}t=-\frac{50}{3}$$

Divide -\frac{25}{3}, the coefficient of the x term, by 2 to get -\frac{25}{6}. Then add the square of -\frac{25}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-\frac{25}{3}t+\left(-\frac{25}{6}\right)^{2}=-\frac{50}{3}+\left(-\frac{25}{6}\right)^{2}$$

Square -\frac{25}{6} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{25}{3}t+\frac{625}{36}=-\frac{50}{3}+\frac{625}{36}$$

Add -\frac{50}{3} to \frac{625}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}-\frac{25}{3}t+\frac{625}{36}=\frac{25}{36}$$

Factor t^{2}-\frac{25}{3}t+\frac{625}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-\frac{25}{6}\right)^{2}=\frac{25}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{25}{6}\right)^{2}}=\sqrt{\frac{25}{36}}$$

Simplify.

$$t-\frac{25}{6}=\frac{5}{6} t-\frac{25}{6}=-\frac{5}{6}$$

Add \frac{25}{6} to both sides of the equation.

$$t=5 t=\frac{10}{3}$$