Solve 10=5t+5t^2

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

Solve for t

$$t=-2$$

$$t=1$$

Steps Using Factoring By Grouping

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

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

Subtract 10 from both sides.

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

Divide both sides by 5.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as t^{2}+at+bt-2. To find a and b, set up a system to be solved.

$$a+b=1 ab=1\left(-2\right)=-2$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1 b=2$$

Rewrite t^{2}+t-2 as \left(t^{2}-t\right)+\left(2t-2\right).

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

Factor out t in the first and 2 in the second group.

$$t\left(t-1\right)+2\left(t-1\right)$$

Factor out common term t-1 by using distributive property.

$$\left(t-1\right)\left(t+2\right)$$

To find equation solutions, solve t-1=0 and t+2=0.

$$t=1 t=-2$$

Steps Using the Quadratic Formula

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

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

Subtract 10 from both sides.

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

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.

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

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

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

Square 5.

$$t=\frac{-5±\sqrt{25-4\times 5\left(-10\right)}}{2\times 5}$$

Multiply -4 times 5.

$$t=\frac{-5±\sqrt{25-20\left(-10\right)}}{2\times 5}$$

Multiply -20 times -10.

$$t=\frac{-5±\sqrt{25+200}}{2\times 5}$$

Add 25 to 200.

$$t=\frac{-5±\sqrt{225}}{2\times 5}$$

Take the square root of 225.

$$t=\frac{-5±15}{2\times 5}$$

Multiply 2 times 5.

$$t=\frac{-5±15}{10}$$

Now solve the equation t=\frac{-5±15}{10} when ± is plus. Add -5 to 15.

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

Divide 10 by 10.

$$t=1$$

Now solve the equation t=\frac{-5±15}{10} when ± is minus. Subtract 15 from -5.

$$t=-\frac{20}{10}$$

Divide -20 by 10.

$$t=-2$$

The equation is now solved.

$$t=1 t=-2$$

Steps for Completing the Square

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

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

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.

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

Divide both sides by 5.

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

Dividing by 5 undoes the multiplication by 5.

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

Divide 5 by 5.

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

Divide 10 by 5.

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

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.

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

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

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

Add 2 to \frac{1}{4}.

$$t^{2}+t+\frac{1}{4}=\frac{9}{4}$$

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}.

$$\left(t+\frac{1}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t+\frac{1}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

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

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

$$t=1 t=-2$$