Solve 0=t^2+5t+4

$$0 = t ^ { 2 } + 5 t + 4$$

Solve for t

$$t=-4$$

$$t=-1$$

Steps Using Factoring

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

$$t^{2}+5t+4=0$$

To solve the equation, factor t^{2}+5t+4 using formula t^{2}+\left(a+b\right)t+ab=\left(t+a\right)\left(t+b\right). To find a and b, set up a system to be solved.

$$a+b=5 ab=4$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 4.

$$1,4 2,2$$

Calculate the sum for each pair.

$$1+4=5 2+2=4$$

The solution is the pair that gives sum 5.

$$a=1 b=4$$

Rewrite factored expression \left(t+a\right)\left(t+b\right) using the obtained values.

$$\left(t+1\right)\left(t+4\right)$$

To find equation solutions, solve t+1=0 and t+4=0.

$$t=-1 t=-4$$

Steps Using Factoring By Grouping

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

$$t^{2}+5t+4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as t^{2}+at+bt+4. To find a and b, set up a system to be solved.

$$a+b=5 ab=1\times 4=4$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 4.

$$1,4 2,2$$

Calculate the sum for each pair.

$$1+4=5 2+2=4$$

The solution is the pair that gives sum 5.

$$a=1 b=4$$

Rewrite t^{2}+5t+4 as \left(t^{2}+t\right)+\left(4t+4\right).

$$\left(t^{2}+t\right)+\left(4t+4\right)$$

Factor out t in the first and 4 in the second group.

$$t\left(t+1\right)+4\left(t+1\right)$$

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

$$\left(t+1\right)\left(t+4\right)$$

To find equation solutions, solve t+1=0 and t+4=0.

$$t=-1 t=-4$$

Steps Using the Quadratic Formula

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

$$t^{2}+5t+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 5 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-5±\sqrt{5^{2}-4\times 4}}{2}$$

Square 5.

$$t=\frac{-5±\sqrt{25-4\times 4}}{2}$$

Multiply -4 times 4.

$$t=\frac{-5±\sqrt{25-16}}{2}$$

Add 25 to -16.

$$t=\frac{-5±\sqrt{9}}{2}$$

Take the square root of 9.

$$t=\frac{-5±3}{2}$$

Now solve the equation t=\frac{-5±3}{2} when ± is plus. Add -5 to 3.

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

Divide -2 by 2.

$$t=-1$$

Now solve the equation t=\frac{-5±3}{2} when ± is minus. Subtract 3 from -5.

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

Divide -8 by 2.

$$t=-4$$

The equation is now solved.

$$t=-1 t=-4$$

Steps for Completing the Square

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

$$t^{2}+5t+4=0$$

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

$$t^{2}+5t=-4$$

Divide 5, the coefficient of the x term, by 2 to get \frac{5}{2}. Then add the square of \frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}+5t+\left(\frac{5}{2}\right)^{2}=-4+\left(\frac{5}{2}\right)^{2}$$

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$t^{2}+5t+\frac{25}{4}=-4+\frac{25}{4}$$

Add -4 to \frac{25}{4}.

$$t^{2}+5t+\frac{25}{4}=\frac{9}{4}$$

Factor t^{2}+5t+\frac{25}{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}.

$$\left(t+\frac{5}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t+\frac{5}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$t+\frac{5}{2}=\frac{3}{2} t+\frac{5}{2}=-\frac{3}{2}$$

Subtract \frac{5}{2} from both sides of the equation.

$$t=-1 t=-4$$