What are the solutions to the equation 40=4t^2+12t ?

The solutions to the equation 40=4t^2+12t are t=2,t=-5

Find the zeros of 40=4t^2+12t

The zeros of 40=4t^2+12t are t=2,t=-5

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40 = 4 t^{2} + 12 t

t = 2, t = - 5

Solution steps

40 = 4 t^{2} + 12 t

Switch sides

4 t^{2} + 12 t = 40

Move

40

to the left side

4 t^{2} + 12 t - 40 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 4 ( - 40 )}{2 \cdot 4}

1 2^{2} - 4 \cdot 4 (- 40) = 28

t_{1, 2} = \frac{- 12 \pm 28}{2 \cdot 4}

Separate the solutions

t_{1} = \frac{- 12 + 28}{2 \cdot 4}, t_{2} = \frac{- 12 - 28}{2 \cdot 4}

t = \frac{- 12 + 28}{2 \cdot 4} : 2

t = \frac{- 12 - 28}{2 \cdot 4} : - 5

The solutions to the quadratic equation are:

t = 2, t = - 5

- 32 c \leq 0

6 x^{2} = 375

\frac{5 a ^{- 5} b ^{4}}{15 a ^{2} b ^{- 6}}

(\frac{1}{81})^{\frac{1}{4}}

5 (x + 2) = 5 x - 2

What are the solutions to the equation 40=4t^2+12t ?
The solutions to the equation 40=4t^2+12t are t=2,t=-5
Find the zeros of 40=4t^2+12t
The zeros of 40=4t^2+12t are t=2,t=-5