Solve 25t-5t^2=20

$$25 t - 5 t ^ { 2 } = 20$$

Solve for t

$$t=1$$

$$t=4$$

Steps Using Factoring By Grouping

Subtract 20 from both sides.

$$25t-5t^{2}-20=0$$

Divide both sides by 5.

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

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

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

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.

$$a+b=5 ab=-\left(-4\right)=4$$

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.

$$a=4 b=1$$

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

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

Factor out -t in -t^{2}+4t.

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

Factor out common term t-4 by using distributive property.

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

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

$$t=4 t=1$$

Steps Using the Quadratic Formula

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.

$$-5t^{2}+25t=20$$

Subtract 20 from both sides of the equation.

$$-5t^{2}+25t-20=20-20$$

Subtracting 20 from itself leaves 0.

$$-5t^{2}+25t-20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 25 for b, and -20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-25±\sqrt{25^{2}-4\left(-5\right)\left(-20\right)}}{2\left(-5\right)}$$

Square 25.

$$t=\frac{-25±\sqrt{625-4\left(-5\right)\left(-20\right)}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$t=\frac{-25±\sqrt{625+20\left(-20\right)}}{2\left(-5\right)}$$

Multiply 20 times -20.

$$t=\frac{-25±\sqrt{625-400}}{2\left(-5\right)}$$

Add 625 to -400.

$$t=\frac{-25±\sqrt{225}}{2\left(-5\right)}$$

Take the square root of 225.

$$t=\frac{-25±15}{2\left(-5\right)}$$

Multiply 2 times -5.

$$t=\frac{-25±15}{-10}$$

Now solve the equation t=\frac{-25±15}{-10} when ± is plus. Add -25 to 15.

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

Divide -10 by -10.

$$t=1$$

Now solve the equation t=\frac{-25±15}{-10} when ± is minus. Subtract 15 from -25.

$$t=-\frac{40}{-10}$$

Divide -40 by -10.

$$t=4$$

The equation is now solved.

$$t=1 t=4$$

Steps for Completing the Square

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.

$$-5t^{2}+25t=20$$

Divide both sides by -5.

$$\frac{-5t^{2}+25t}{-5}=\frac{20}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$t^{2}+\frac{25}{-5}t=\frac{20}{-5}$$

Divide 25 by -5.

$$t^{2}-5t=\frac{20}{-5}$$

Divide 20 by -5.

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

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.

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

Add \frac{5}{2} to both sides of the equation.

$$t=4 t=1$$