What are the solutions to the equation 40=2t+2t^2 ?

The solutions to the equation 40=2t+2t^2 are t=4,t=-5

Find the zeros of 40=2t+2t^2

The zeros of 40=2t+2t^2 are t=4,t=-5

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40 = 2 t + 2 t^{2}

t = 4, t = - 5

Solution steps

40 = 2 t + 2 t^{2}

Switch sides

2 t + 2 t^{2} = 40

Move

40

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

2^{2} - 4 \cdot 2 (- 40) = 18

t_{1, 2} = \frac{- 2 \pm 18}{2 \cdot 2}

Separate the solutions

t_{1} = \frac{- 2 + 18}{2 \cdot 2}, t_{2} = \frac{- 2 - 18}{2 \cdot 2}

t = \frac{- 2 + 18}{2 \cdot 2} : 4

t = \frac{- 2 - 18}{2 \cdot 2} : - 5

The solutions to the quadratic equation are:

t = 4, t = - 5

x^{2} - 7 x - \frac{3}{4} = 0

n^{2} + n - 12 = 0

x - 6 = - 3 x^{2}

t^{2} + 22 t - 5 = 0

(x + 4)^{2} = 12

What are the solutions to the equation 40=2t+2t^2 ?
The solutions to the equation 40=2t+2t^2 are t=4,t=-5
Find the zeros of 40=2t+2t^2
The zeros of 40=2t+2t^2 are t=4,t=-5