Solve 5(t^2+2t-4)=0

$$5 ( t ^ { 2 } + 2 t - 4 ) = 0$$

Solve for t

$$t=\sqrt{5}-1\approx 1.236067977$$

$$t=-\sqrt{5}-1\approx -3.236067977$$

Steps Using the Quadratic Formula

Divide both sides by 5. Zero divided by any non-zero number gives zero.

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

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

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

Square 2.

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

Multiply -4 times -4.

$$t=\frac{-2±\sqrt{4+16}}{2}$$

Add 4 to 16.

$$t=\frac{-2±\sqrt{20}}{2}$$

Take the square root of 20.

$$t=\frac{-2±2\sqrt{5}}{2}$$

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

$$t=\frac{2\sqrt{5}-2}{2}$$

Divide -2+2\sqrt{5} by 2.

$$t=\sqrt{5}-1$$

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

$$t=\frac{-2\sqrt{5}-2}{2}$$

Divide -2-2\sqrt{5} by 2.

$$t=-\sqrt{5}-1$$

The equation is now solved.

$$t=\sqrt{5}-1 t=-\sqrt{5}-1$$

Steps for Completing the Square

Divide both sides by 5. Zero divided by any non-zero number gives zero.

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

Add 4 to both sides. Anything plus zero gives itself.

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

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.

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

Square 1.

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

Add 4 to 1.

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

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}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(t+1\right)^{2}}=\sqrt{5}$$

Simplify.

$$t+1=\sqrt{5} t+1=-\sqrt{5}$$

Subtract 1 from both sides of the equation.

$$t=\sqrt{5}-1 t=-\sqrt{5}-1$$