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

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

Solve for t

$$t = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

$$t=\frac{1}{2}=0.5$$

Steps Using the Quadratic Formula

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

$$t^{2}+t^{2}+2t+1=\frac{5}{2}$$

Combine t^{2} and t^{2} to get 2t^{2}.

$$2t^{2}+2t+1=\frac{5}{2}$$

Subtract \frac{5}{2} from both sides.

$$2t^{2}+2t+1-\frac{5}{2}=0$$

Subtract \frac{5}{2} from 1 to get -\frac{3}{2}.

$$2t^{2}+2t-\frac{3}{2}=0$$

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

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

Square 2.

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

Multiply -4 times 2.

$$t=\frac{-2±\sqrt{4-8\left(-\frac{3}{2}\right)}}{2\times 2}$$

Multiply -8 times -\frac{3}{2}.

$$t=\frac{-2±\sqrt{4+12}}{2\times 2}$$

Add 4 to 12.

$$t=\frac{-2±\sqrt{16}}{2\times 2}$$

Take the square root of 16.

$$t=\frac{-2±4}{2\times 2}$$

Multiply 2 times 2.

$$t=\frac{-2±4}{4}$$

Now solve the equation t=\frac{-2±4}{4} when ± is plus. Add -2 to 4.

$$t=\frac{2}{4}$$

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

$$t=\frac{1}{2}$$

Now solve the equation t=\frac{-2±4}{4} when ± is minus. Subtract 4 from -2.

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

Reduce the fraction \frac{-6}{4} to lowest terms by extracting and canceling out 2.

$$t=-\frac{3}{2}$$

The equation is now solved.

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

Steps for Completing the Square

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

$$t^{2}+t^{2}+2t+1=\frac{5}{2}$$

Combine t^{2} and t^{2} to get 2t^{2}.

$$2t^{2}+2t+1=\frac{5}{2}$$

Subtract 1 from both sides.

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

Subtract 1 from \frac{5}{2} to get \frac{3}{2}.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide 2 by 2.

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

Divide \frac{3}{2} by 2.

$$t^{2}+t=\frac{3}{4}$$

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

$$t^{2}+t+\left(\frac{1}{2}\right)^{2}=\frac{3}{4}+\left(\frac{1}{2}\right)^{2}$$

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$t^{2}+t+\frac{1}{4}=\frac{3+1}{4}$$

Add \frac{3}{4} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}+t+\frac{1}{4}=1$$

Factor t^{2}+t+\frac{1}{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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$t+\frac{1}{2}=1 t+\frac{1}{2}=-1$$

Subtract \frac{1}{2} from both sides of the equation.

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