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

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

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

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

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

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

Decimal Notation

t = - 3.37228 \dots, t = 2.37228 \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{33 - 1}{2}

The solutions to the quadratic equation are:

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

(6 x + 4)^{2} = 16

x, 9 x^{2} + 12 x y + 4 y^{2} - 18 x - 16 y - 11 = 0

16 x^{2} - 235 x + 151 = 0

3 x^{2} + 1 = - 13 x - 3

[\frac{7 x ( x )}{2 + ( - 7 )}] = 21

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