What are the solutions to the equation 2.4+t+0.1t^2=0 ?

The solutions to the equation 2.4+t+0.1t^2=0 are t=-4,t=-6

Find the zeros of 2.4+t+0.1t^2=0

The zeros of 2.4+t+0.1t^2=0 are t=-4,t=-6

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2.4 + t + 0.1 t^{2} = 0

t = - 4, t = - 6

Solution steps

2.4 + t + 0.1 t^{2} = 0

Multiply both sides by

10

24 + 10 t + t^{2} = 0

Write in the standard form

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

t^{2} + 10 t + 24 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 1 \cdot 24}{2 \cdot 1}

1 0^{2} - 4 \cdot 1 \cdot 24 = 2

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

Separate the solutions

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

t = \frac{- 10 + 2}{2 \cdot 1} : - 4

t = \frac{- 10 - 2}{2 \cdot 1} : - 6

The solutions to the quadratic equation are:

t = - 4, t = - 6

5 (x^{2} - 1) = 4 (x^{2} + 1)

48 x^{2} + 12 x - 90 = 0

x^{2} - 4 x + 2 = 10

x, 4 x^{2} + 4 y^{2} = 0

a, x^{2} + 2 a x + a^{2} = \frac{( x \cdot y )}{p}

What are the solutions to the equation 2.4+t+0.1t^2=0 ?
The solutions to the equation 2.4+t+0.1t^2=0 are t=-4,t=-6
Find the zeros of 2.4+t+0.1t^2=0
The zeros of 2.4+t+0.1t^2=0 are t=-4,t=-6