Solve 5-2+-3/3=(t+1)^2-3++1/3

$$5 - \frac { 2 + - 3 } { 3 } = ( t + 1 ) ^ { 2 } - \frac { 3 + + 1 } { 3 }$$

Solve for t

$$t=\frac{2\sqrt{15}}{3}-1\approx 1.581988897$$

$$t=-\frac{2\sqrt{15}}{3}-1\approx -3.581988897$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 3.

$$15-\left(2-3\right)=3\left(t+1\right)^{2}-\left(3+1\right)$$

Subtract 3 from 2 to get -1.

$$15-\left(-1\right)=3\left(t+1\right)^{2}-\left(3+1\right)$$

The opposite of -1 is 1.

$$15+1=3\left(t+1\right)^{2}-\left(3+1\right)$$

Add 15 and 1 to get 16.

$$16=3\left(t+1\right)^{2}-\left(3+1\right)$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(t+1\right)^{2}.

$$16=3\left(t^{2}+2t+1\right)-\left(3+1\right)$$

Use the distributive property to multiply 3 by t^{2}+2t+1.

$$16=3t^{2}+6t+3-\left(3+1\right)$$

Add 3 and 1 to get 4.

$$16=3t^{2}+6t+3-4$$

Subtract 4 from 3 to get -1.

$$16=3t^{2}+6t-1$$

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

$$3t^{2}+6t-1=16$$

Subtract 16 from both sides.

$$3t^{2}+6t-1-16=0$$

Subtract 16 from -1 to get -17.

$$3t^{2}+6t-17=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 6 for b, and -17 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-6±\sqrt{6^{2}-4\times 3\left(-17\right)}}{2\times 3}$$

Square 6.

$$t=\frac{-6±\sqrt{36-4\times 3\left(-17\right)}}{2\times 3}$$

Multiply -4 times 3.

$$t=\frac{-6±\sqrt{36-12\left(-17\right)}}{2\times 3}$$

Multiply -12 times -17.

$$t=\frac{-6±\sqrt{36+204}}{2\times 3}$$

Add 36 to 204.

$$t=\frac{-6±\sqrt{240}}{2\times 3}$$

Take the square root of 240.

$$t=\frac{-6±4\sqrt{15}}{2\times 3}$$

Multiply 2 times 3.

$$t=\frac{-6±4\sqrt{15}}{6}$$

Now solve the equation t=\frac{-6±4\sqrt{15}}{6} when ± is plus. Add -6 to 4\sqrt{15}.

$$t=\frac{4\sqrt{15}-6}{6}$$

Divide -6+4\sqrt{15} by 6.

$$t=\frac{2\sqrt{15}}{3}-1$$

Now solve the equation t=\frac{-6±4\sqrt{15}}{6} when ± is minus. Subtract 4\sqrt{15} from -6.

$$t=\frac{-4\sqrt{15}-6}{6}$$

Divide -6-4\sqrt{15} by 6.

$$t=-\frac{2\sqrt{15}}{3}-1$$

The equation is now solved.

$$t=\frac{2\sqrt{15}}{3}-1 t=-\frac{2\sqrt{15}}{3}-1$$

Steps for Completing the Square

Multiply both sides of the equation by 3.

$$15-\left(2-3\right)=3\left(t+1\right)^{2}-\left(3+1\right)$$

Subtract 3 from 2 to get -1.

$$15-\left(-1\right)=3\left(t+1\right)^{2}-\left(3+1\right)$$

The opposite of -1 is 1.

$$15+1=3\left(t+1\right)^{2}-\left(3+1\right)$$

Add 15 and 1 to get 16.

$$16=3\left(t+1\right)^{2}-\left(3+1\right)$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(t+1\right)^{2}.

$$16=3\left(t^{2}+2t+1\right)-\left(3+1\right)$$

Use the distributive property to multiply 3 by t^{2}+2t+1.

$$16=3t^{2}+6t+3-\left(3+1\right)$$

Add 3 and 1 to get 4.

$$16=3t^{2}+6t+3-4$$

Subtract 4 from 3 to get -1.

$$16=3t^{2}+6t-1$$

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

$$3t^{2}+6t-1=16$$

Add 1 to both sides.

$$3t^{2}+6t=16+1$$

Add 16 and 1 to get 17.

$$3t^{2}+6t=17$$

Divide both sides by 3.

$$\frac{3t^{2}+6t}{3}=\frac{17}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$t^{2}+\frac{6}{3}t=\frac{17}{3}$$

Divide 6 by 3.

$$t^{2}+2t=\frac{17}{3}$$

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}+2t+1^{2}=\frac{17}{3}+1^{2}$$

Square 1.

$$t^{2}+2t+1=\frac{17}{3}+1$$

Add \frac{17}{3} to 1.

$$t^{2}+2t+1=\frac{20}{3}$$

Factor t^{2}+2t+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t+1\right)^{2}=\frac{20}{3}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t+1\right)^{2}}=\sqrt{\frac{20}{3}}$$

Simplify.

$$t+1=\frac{2\sqrt{15}}{3} t+1=-\frac{2\sqrt{15}}{3}$$

Subtract 1 from both sides of the equation.

$$t=\frac{2\sqrt{15}}{3}-1 t=-\frac{2\sqrt{15}}{3}-1$$