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-5t^{2}+4t=9
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}+4t-9=9-9
Subtract 9 from both sides of the equation.
-5t^{2}+4t-9=0
Subtracting 9 from itself leaves 0.
t=\frac{-4±\sqrt{4^{2}-4\left(-5\right)\left(-9\right)}}{2\left(-5\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 4 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
t=\frac{-4±\sqrt{16-4\left(-5\right)\left(-9\right)}}{2\left(-5\right)}
Square 4.
t=\frac{-4±\sqrt{16+20\left(-9\right)}}{2\left(-5\right)}
Multiply -4 times -5.
t=\frac{-4±\sqrt{16-180}}{2\left(-5\right)}
Multiply 20 times -9.
t=\frac{-4±\sqrt{-164}}{2\left(-5\right)}
Add 16 to -180.
t=\frac{-4±2\sqrt{41}i}{2\left(-5\right)}
Take the square root of -164.
t=\frac{-4±2\sqrt{41}i}{-10}
Multiply 2 times -5.
t=\frac{-4+2\sqrt{41}i}{-10}
Now solve the equation t=\frac{-4±2\sqrt{41}i}{-10} when ± is plus. Add -4 to 2i\sqrt{41}.
t=\frac{-\sqrt{41}i+2}{5}
Divide -4+2i\sqrt{41} by -10.
t=\frac{-2\sqrt{41}i-4}{-10}
Now solve the equation t=\frac{-4±2\sqrt{41}i}{-10} when ± is minus. Subtract 2i\sqrt{41} from -4.
t=\frac{2+\sqrt{41}i}{5}
Divide -4-2i\sqrt{41} by -10.
t=\frac{-\sqrt{41}i+2}{5} t=\frac{2+\sqrt{41}i}{5}
The equation is now solved.
-5t^{2}+4t=9
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.
\frac{-5t^{2}+4t}{-5}=\frac{9}{-5}
Divide both sides by -5.
t^{2}+\frac{4}{-5}t=\frac{9}{-5}
Dividing by -5 undoes the multiplication by -5.
t^{2}-\frac{4}{5}t=\frac{9}{-5}
Divide 4 by -5.
t^{2}-\frac{4}{5}t=-\frac{9}{5}
Divide 9 by -5.
t^{2}-\frac{4}{5}t+\left(-\frac{2}{5}\right)^{2}=-\frac{9}{5}+\left(-\frac{2}{5}\right)^{2}
Divide -\frac{4}{5}, the coefficient of the x term, by 2 to get -\frac{2}{5}. Then add the square of -\frac{2}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
t^{2}-\frac{4}{5}t+\frac{4}{25}=-\frac{9}{5}+\frac{4}{25}
Square -\frac{2}{5} by squaring both the numerator and the denominator of the fraction.
t^{2}-\frac{4}{5}t+\frac{4}{25}=-\frac{41}{25}
Add -\frac{9}{5} to \frac{4}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(t-\frac{2}{5}\right)^{2}=-\frac{41}{25}
Factor t^{2}-\frac{4}{5}t+\frac{4}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(t-\frac{2}{5}\right)^{2}}=\sqrt{-\frac{41}{25}}
Take the square root of both sides of the equation.
t-\frac{2}{5}=\frac{\sqrt{41}i}{5} t-\frac{2}{5}=-\frac{\sqrt{41}i}{5}
Simplify.
t=\frac{2+\sqrt{41}i}{5} t=\frac{-\sqrt{41}i+2}{5}
Add \frac{2}{5} to both sides of the equation.