What are the solutions to the equation 100=15t+(4t^2)/2 ?

The solutions to the equation 100=15t+(4t^2)/2 are t=(5(sqrt(41)-3))/4 ,t=-(5(3+sqrt(41)))/4

Find the zeros of 100=15t+(4t^2)/2

The zeros of 100=15t+(4t^2)/2 are t=(5(sqrt(41)-3))/4 ,t=-(5(3+sqrt(41)))/4

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100 = 15 t + \frac{4 t ^{2}}{2}

t = \frac{5 ( 41 - 3 )}{4}, t = - \frac{5 ( 3 + 41 )}{4}

Solution steps

100 = 15 t + \frac{4 t ^{2}}{2}

Multiply both sides by

2

200 = 30 t + 4 t^{2}

Switch sides

30 t + 4 t^{2} = 200

Move

200

to the left side

30 t + 4 t^{2} - 200 = 0

Write in the standard form

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

4 t^{2} + 30 t - 200 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 30 \pm 3 0 ^{2} - 4 \cdot 4 ( - 200 )}{2 \cdot 4}

3 0^{2} - 4 \cdot 4 (- 200) = 1041

t_{1, 2} = \frac{- 30 \pm 10 41}{2 \cdot 4}

Separate the solutions

t_{1} = \frac{- 30 + 10 41}{2 \cdot 4}, t_{2} = \frac{- 30 - 10 41}{2 \cdot 4}

t = \frac{- 30 + 10 41}{2 \cdot 4} : \frac{5 ( 41 - 3 )}{4}

t = \frac{- 30 - 10 41}{2 \cdot 4} : - \frac{5 ( 3 + 41 )}{4}

The solutions to the quadratic equation are:

t = \frac{5 ( 41 - 3 )}{4}, t = - \frac{5 ( 3 + 41 )}{4}

- x^{2} - 3 x - 1

2 (x - 3)^{2} = 50

3 + 5 x - 2 x^{2} = 0

4 x^{2} + 2 x = 13

11 x^{2} - 4 x + 11 = 0

What are the solutions to the equation 100=15t+(4t^2)/2 ?
The solutions to the equation 100=15t+(4t^2)/2 are t=(5(sqrt(41)-3))/4 ,t=-(5(3+sqrt(41)))/4
Find the zeros of 100=15t+(4t^2)/2
The zeros of 100=15t+(4t^2)/2 are t=(5(sqrt(41)-3))/4 ,t=-(5(3+sqrt(41)))/4