Solve 5=10t+5t^2

$$5 = 10 t + 5 t ^ { 2 }$$

Solve for t

$$t=\sqrt{2}-1\approx 0.414213562$$

$$t=-\sqrt{2}-1\approx -2.414213562$$

Steps Using the Quadratic Formula

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

$$10t+5t^{2}=5$$

Subtract 5 from both sides.

$$10t+5t^{2}-5=0$$

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.

$$5t^{2}+10t-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 10 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-10±\sqrt{10^{2}-4\times 5\left(-5\right)}}{2\times 5}$$

Square 10.

$$t=\frac{-10±\sqrt{100-4\times 5\left(-5\right)}}{2\times 5}$$

Multiply -4 times 5.

$$t=\frac{-10±\sqrt{100-20\left(-5\right)}}{2\times 5}$$

Multiply -20 times -5.

$$t=\frac{-10±\sqrt{100+100}}{2\times 5}$$

Add 100 to 100.

$$t=\frac{-10±\sqrt{200}}{2\times 5}$$

Take the square root of 200.

$$t=\frac{-10±10\sqrt{2}}{2\times 5}$$

Multiply 2 times 5.

$$t=\frac{-10±10\sqrt{2}}{10}$$

Now solve the equation t=\frac{-10±10\sqrt{2}}{10} when ± is plus. Add -10 to 10\sqrt{2}.

$$t=\frac{10\sqrt{2}-10}{10}$$

Divide -10+10\sqrt{2} by 10.

$$t=\sqrt{2}-1$$

Now solve the equation t=\frac{-10±10\sqrt{2}}{10} when ± is minus. Subtract 10\sqrt{2} from -10.

$$t=\frac{-10\sqrt{2}-10}{10}$$

Divide -10-10\sqrt{2} by 10.

$$t=-\sqrt{2}-1$$

The equation is now solved.

$$t=\sqrt{2}-1 t=-\sqrt{2}-1$$

Steps for Completing the Square

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

$$10t+5t^{2}=5$$

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.

$$5t^{2}+10t=5$$

Divide both sides by 5.

$$\frac{5t^{2}+10t}{5}=\frac{5}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$t^{2}+\frac{10}{5}t=\frac{5}{5}$$

Divide 10 by 5.

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

Divide 5 by 5.

$$t^{2}+2t=1$$

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.

$$t^{2}+2t+1^{2}=1+1^{2}$$

Square 1.

$$t^{2}+2t+1=1+1$$

Add 1 to 1.

$$t^{2}+2t+1=2$$

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract 1 from both sides of the equation.

$$t=\sqrt{2}-1 t=-\sqrt{2}-1$$