Solve 0=-5t^2+23t+10

$$0 = - 5 t ^ { 2 } + 23 t + 10$$

Solve for t

$$t=5$$

$$t=-\frac{2}{5}=-0.4$$

Steps Using Factoring By Grouping

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

$$-5t^{2}+23t+10=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -5t^{2}+at+bt+10. To find a and b, set up a system to be solved.

$$a+b=23 ab=-5\times 10=-50$$

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. List all such integer pairs that give product -50.

$$-1,50 -2,25 -5,10$$

Calculate the sum for each pair.

$$-1+50=49 -2+25=23 -5+10=5$$

The solution is the pair that gives sum 23.

$$a=25 b=-2$$

Rewrite -5t^{2}+23t+10 as \left(-5t^{2}+25t\right)+\left(-2t+10\right).

$$\left(-5t^{2}+25t\right)+\left(-2t+10\right)$$

Factor out 5t in the first and 2 in the second group.

$$5t\left(-t+5\right)+2\left(-t+5\right)$$

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

$$\left(-t+5\right)\left(5t+2\right)$$

To find equation solutions, solve -t+5=0 and 5t+2=0.

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

Steps Using the Quadratic Formula

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

$$-5t^{2}+23t+10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 23 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-23±\sqrt{23^{2}-4\left(-5\right)\times 10}}{2\left(-5\right)}$$

Square 23.

$$t=\frac{-23±\sqrt{529-4\left(-5\right)\times 10}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$t=\frac{-23±\sqrt{529+20\times 10}}{2\left(-5\right)}$$

Multiply 20 times 10.

$$t=\frac{-23±\sqrt{529+200}}{2\left(-5\right)}$$

Add 529 to 200.

$$t=\frac{-23±\sqrt{729}}{2\left(-5\right)}$$

Take the square root of 729.

$$t=\frac{-23±27}{2\left(-5\right)}$$

Multiply 2 times -5.

$$t=\frac{-23±27}{-10}$$

Now solve the equation t=\frac{-23±27}{-10} when ± is plus. Add -23 to 27.

$$t=\frac{4}{-10}$$

Reduce the fraction \frac{4}{-10} to lowest terms by extracting and canceling out 2.

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

Now solve the equation t=\frac{-23±27}{-10} when ± is minus. Subtract 27 from -23.

$$t=-\frac{50}{-10}$$

Divide -50 by -10.

$$t=5$$

The equation is now solved.

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

Steps for Completing the Square

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

$$-5t^{2}+23t+10=0$$

Subtract 10 from both sides. Anything subtracted from zero gives its negation.

$$-5t^{2}+23t=-10$$

Divide both sides by -5.

$$\frac{-5t^{2}+23t}{-5}=-\frac{10}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$t^{2}+\frac{23}{-5}t=-\frac{10}{-5}$$

Divide 23 by -5.

$$t^{2}-\frac{23}{5}t=-\frac{10}{-5}$$

Divide -10 by -5.

$$t^{2}-\frac{23}{5}t=2$$

Divide -\frac{23}{5}, the coefficient of the x term, by 2 to get -\frac{23}{10}. Then add the square of -\frac{23}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-\frac{23}{5}t+\left(-\frac{23}{10}\right)^{2}=2+\left(-\frac{23}{10}\right)^{2}$$

Square -\frac{23}{10} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{23}{5}t+\frac{529}{100}=2+\frac{529}{100}$$

Add 2 to \frac{529}{100}.

$$t^{2}-\frac{23}{5}t+\frac{529}{100}=\frac{729}{100}$$

Factor t^{2}-\frac{23}{5}t+\frac{529}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-\frac{23}{10}\right)^{2}=\frac{729}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{23}{10}\right)^{2}}=\sqrt{\frac{729}{100}}$$

Simplify.

$$t-\frac{23}{10}=\frac{27}{10} t-\frac{23}{10}=-\frac{27}{10}$$

Add \frac{23}{10} to both sides of the equation.

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