Solve 2.5=4-left(4-tright)^2

$$2.5 = 4- { \left(4-t \right) }^{ 2 }$$

Solve for t

$$t=-\frac{\sqrt{6}}{2}+4\approx 2.775255129$$

$$t=\frac{\sqrt{6}}{2}+4\approx 5.224744871$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-t\right)^{2}.

$$2.5=4-\left(16-8t+t^{2}\right)$$

To find the opposite of 16-8t+t^{2}, find the opposite of each term.

$$2.5=4-16+8t-t^{2}$$

Subtract 16 from 4 to get -12.

$$2.5=-12+8t-t^{2}$$

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

$$-12+8t-t^{2}=2.5$$

Subtract 2.5 from both sides.

$$-12+8t-t^{2}-2.5=0$$

Subtract 2.5 from -12 to get -14.5.

$$-14.5+8t-t^{2}=0$$

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.

$$-t^{2}+8t-14.5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 8 for b, and -14.5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-8±\sqrt{8^{2}-4\left(-1\right)\left(-14.5\right)}}{2\left(-1\right)}$$

Square 8.

$$t=\frac{-8±\sqrt{64-4\left(-1\right)\left(-14.5\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$t=\frac{-8±\sqrt{64+4\left(-14.5\right)}}{2\left(-1\right)}$$

Multiply 4 times -14.5.

$$t=\frac{-8±\sqrt{64-58}}{2\left(-1\right)}$$

Add 64 to -58.

$$t=\frac{-8±\sqrt{6}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$t=\frac{-8±\sqrt{6}}{-2}$$

Now solve the equation t=\frac{-8±\sqrt{6}}{-2} when ± is plus. Add -8 to \sqrt{6}.

$$t=\frac{\sqrt{6}-8}{-2}$$

Divide -8+\sqrt{6} by -2.

$$t=-\frac{\sqrt{6}}{2}+4$$

Now solve the equation t=\frac{-8±\sqrt{6}}{-2} when ± is minus. Subtract \sqrt{6} from -8.

$$t=\frac{-\sqrt{6}-8}{-2}$$

Divide -8-\sqrt{6} by -2.

$$t=\frac{\sqrt{6}}{2}+4$$

The equation is now solved.

$$t=-\frac{\sqrt{6}}{2}+4 t=\frac{\sqrt{6}}{2}+4$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-t\right)^{2}.

$$2.5=4-\left(16-8t+t^{2}\right)$$

To find the opposite of 16-8t+t^{2}, find the opposite of each term.

$$2.5=4-16+8t-t^{2}$$

Subtract 16 from 4 to get -12.

$$2.5=-12+8t-t^{2}$$

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

$$-12+8t-t^{2}=2.5$$

Add 12 to both sides.

$$8t-t^{2}=2.5+12$$

Add 2.5 and 12 to get 14.5.

$$8t-t^{2}=14.5$$

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.

$$-t^{2}+8t=14.5$$

Divide both sides by -1.

$$\frac{-t^{2}+8t}{-1}=\frac{14.5}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$t^{2}+\frac{8}{-1}t=\frac{14.5}{-1}$$

Divide 8 by -1.

$$t^{2}-8t=\frac{14.5}{-1}$$

Divide 14.5 by -1.

$$t^{2}-8t=-14.5$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-8t+\left(-4\right)^{2}=-14.5+\left(-4\right)^{2}$$

Square -4.

$$t^{2}-8t+16=-14.5+16$$

Add -14.5 to 16.

$$t^{2}-8t+16=1.5$$

Factor t^{2}-8t+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-4\right)^{2}=1.5$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-4\right)^{2}}=\sqrt{1.5}$$

Simplify.

$$t-4=\frac{\sqrt{6}}{2} t-4=-\frac{\sqrt{6}}{2}$$

Add 4 to both sides of the equation.

$$t=\frac{\sqrt{6}}{2}+4 t=-\frac{\sqrt{6}}{2}+4$$