What are the solutions to the equation 0=40+5*t-5*t^2 ?

The solutions to the equation 0=40+5*t-5*t^2 are t=-(-1+sqrt(33))/2 ,t=(1+sqrt(33))/2

Find the zeros of 0=40+5*t-5*t^2

The zeros of 0=40+5*t-5*t^2 are t=-(-1+sqrt(33))/2 ,t=(1+sqrt(33))/2

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0 = 40 + 5 \cdot t - 5 \cdot t^{2}

t = - \frac{- 1 + 33}{2}, t = \frac{1 + 33}{2}

Decimal Notation

t = - 2.37228 \dots, t = 3.37228 \dots

Solution steps

0 = 40 + 5 t - 5 t^{2}

Switch sides

40 + 5 t - 5 t^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

- 5 t^{2} + 5 t + 40 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 ( - 5 ) \cdot 40}{2 ( - 5 )}

5^{2} - 4 (- 5) \cdot 40 = 533

t_{1, 2} = \frac{- 5 \pm 5 33}{2 ( - 5 )}

Separate the solutions

t_{1} = \frac{- 5 + 5 33}{2 ( - 5 )}, t_{2} = \frac{- 5 - 5 33}{2 ( - 5 )}

t = \frac{- 5 + 5 33}{2 ( - 5 )} : - \frac{- 1 + 33}{2}

t = \frac{- 5 - 5 33}{2 ( - 5 )} : \frac{1 + 33}{2}

The solutions to the quadratic equation are:

t = - \frac{- 1 + 33}{2}, t = \frac{1 + 33}{2}

p^{2} + 10 \cdot 10 = 4160

4 X^{2} - 32 X + 20 = 0

14 x^{2} - 45 x - 14 = 0

- 4 x^{2} - 24 x - 28 = 0

12 x^{2} - 7 x - 1 = 0

What are the solutions to the equation 0=40+5*t-5*t^2 ?
The solutions to the equation 0=40+5*t-5*t^2 are t=-(-1+sqrt(33))/2 ,t=(1+sqrt(33))/2
Find the zeros of 0=40+5*t-5*t^2
The zeros of 0=40+5*t-5*t^2 are t=-(-1+sqrt(33))/2 ,t=(1+sqrt(33))/2