Solve 6=-16t^2+24t+1

$$6 = - 16 t ^ { 2 } + 24 t + 1$$

Solve for t

$$t=\frac{1}{4}=0.25$$

$$t = \frac{5}{4} = 1\frac{1}{4} = 1.25$$

Steps Using Factoring By Grouping

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

$$-16t^{2}+24t+1=6$$

Subtract 6 from both sides.

$$-16t^{2}+24t+1-6=0$$

Subtract 6 from 1 to get -5.

$$-16t^{2}+24t-5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -16t^{2}+at+bt-5. To find a and b, set up a system to be solved.

$$a+b=24 ab=-16\left(-5\right)=80$$

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 80.

$$1,80 2,40 4,20 5,16 8,10$$

Calculate the sum for each pair.

$$1+80=81 2+40=42 4+20=24 5+16=21 8+10=18$$

The solution is the pair that gives sum 24.

$$a=20 b=4$$

Rewrite -16t^{2}+24t-5 as \left(-16t^{2}+20t\right)+\left(4t-5\right).

$$\left(-16t^{2}+20t\right)+\left(4t-5\right)$$

Factor out -4t in -16t^{2}+20t.

$$-4t\left(4t-5\right)+4t-5$$

Factor out common term 4t-5 by using distributive property.

$$\left(4t-5\right)\left(-4t+1\right)$$

To find equation solutions, solve 4t-5=0 and -4t+1=0.

$$t=\frac{5}{4} t=\frac{1}{4}$$

Steps Using the Quadratic Formula

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

$$-16t^{2}+24t+1=6$$

Subtract 6 from both sides.

$$-16t^{2}+24t+1-6=0$$

Subtract 6 from 1 to get -5.

$$-16t^{2}+24t-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -16 for a, 24 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-24±\sqrt{24^{2}-4\left(-16\right)\left(-5\right)}}{2\left(-16\right)}$$

Square 24.

$$t=\frac{-24±\sqrt{576-4\left(-16\right)\left(-5\right)}}{2\left(-16\right)}$$

Multiply -4 times -16.

$$t=\frac{-24±\sqrt{576+64\left(-5\right)}}{2\left(-16\right)}$$

Multiply 64 times -5.

$$t=\frac{-24±\sqrt{576-320}}{2\left(-16\right)}$$

Add 576 to -320.

$$t=\frac{-24±\sqrt{256}}{2\left(-16\right)}$$

Take the square root of 256.

$$t=\frac{-24±16}{2\left(-16\right)}$$

Multiply 2 times -16.

$$t=\frac{-24±16}{-32}$$

Now solve the equation t=\frac{-24±16}{-32} when ± is plus. Add -24 to 16.

$$t=-\frac{8}{-32}$$

Reduce the fraction \frac{-8}{-32} to lowest terms by extracting and canceling out 8.

$$t=\frac{1}{4}$$

Now solve the equation t=\frac{-24±16}{-32} when ± is minus. Subtract 16 from -24.

$$t=-\frac{40}{-32}$$

Reduce the fraction \frac{-40}{-32} to lowest terms by extracting and canceling out 8.

$$t=\frac{5}{4}$$

The equation is now solved.

$$t=\frac{1}{4} t=\frac{5}{4}$$

Steps for Completing the Square

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

$$-16t^{2}+24t+1=6$$

Subtract 1 from both sides.

$$-16t^{2}+24t=6-1$$

Subtract 1 from 6 to get 5.

$$-16t^{2}+24t=5$$

Divide both sides by -16.

$$\frac{-16t^{2}+24t}{-16}=\frac{5}{-16}$$

Dividing by -16 undoes the multiplication by -16.

$$t^{2}+\frac{24}{-16}t=\frac{5}{-16}$$

Reduce the fraction \frac{24}{-16} to lowest terms by extracting and canceling out 8.

$$t^{2}-\frac{3}{2}t=\frac{5}{-16}$$

Divide 5 by -16.

$$t^{2}-\frac{3}{2}t=-\frac{5}{16}$$

Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-\frac{3}{2}t+\left(-\frac{3}{4}\right)^{2}=-\frac{5}{16}+\left(-\frac{3}{4}\right)^{2}$$

Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{3}{2}t+\frac{9}{16}=\frac{-5+9}{16}$$

Add -\frac{5}{16} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}-\frac{3}{2}t+\frac{9}{16}=\frac{1}{4}$$

Factor t^{2}-\frac{3}{2}t+\frac{9}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-\frac{3}{4}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{3}{4}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$t-\frac{3}{4}=\frac{1}{2} t-\frac{3}{4}=-\frac{1}{2}$$

Add \frac{3}{4} to both sides of the equation.

$$t=\frac{5}{4} t=\frac{1}{4}$$