What are the solutions to the equation 0.15t^2=150-4t ?

The solutions to the equation 0.15t^2=150-4t are t=(10(sqrt(106)-4))/3 ,t=-(10(4+sqrt(106)))/3

Find the zeros of 0.15t^2=150-4t

The zeros of 0.15t^2=150-4t are t=(10(sqrt(106)-4))/3 ,t=-(10(4+sqrt(106)))/3

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0.15 t^{2} = 150 - 4 t

t = \frac{10 ( 106 - 4 )}{3}, t = - \frac{10 ( 4 + 106 )}{3}

Decimal Notation

t = 20.98543 \dots, t = - 47.65210 \dots

Solution steps

0.15 t^{2} = 150 - 4 t

Multiply both sides by

100

15 t^{2} = 15000 - 400 t

Move

400 t

to the left side

15 t^{2} + 400 t = 15000

Move

15000

to the left side

15 t^{2} + 400 t - 15000 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 400 \pm 40 0 ^{2} - 4 \cdot 15 ( - 15000 )}{2 \cdot 15}

40 0^{2} - 4 \cdot 15 (- 15000) = 100106

t_{1, 2} = \frac{- 400 \pm 100 106}{2 \cdot 15}

Separate the solutions

t_{1} = \frac{- 400 + 100 106}{2 \cdot 15}, t_{2} = \frac{- 400 - 100 106}{2 \cdot 15}

t = \frac{- 400 + 100 106}{2 \cdot 15} : \frac{10 ( 106 - 4 )}{3}

t = \frac{- 400 - 100 106}{2 \cdot 15} : - \frac{10 ( 4 + 106 )}{3}

The solutions to the quadratic equation are:

t = \frac{10 ( 106 - 4 )}{3}, t = - \frac{10 ( 4 + 106 )}{3}

7 (x + 2) = 42

3^{5}

lo g_{2} (32) = x

x^{3} - 5 x^{2} + 8 x - 4

\frac{4}{5 - 7 i}

What are the solutions to the equation 0.15t^2=150-4t ?
The solutions to the equation 0.15t^2=150-4t are t=(10(sqrt(106)-4))/3 ,t=-(10(4+sqrt(106)))/3
Find the zeros of 0.15t^2=150-4t
The zeros of 0.15t^2=150-4t are t=(10(sqrt(106)-4))/3 ,t=-(10(4+sqrt(106)))/3