What are the solutions to the equation -4t^2+32t=-132 ?

The solutions to the equation -4t^2+32t=-132 are t=-3,t=11

Find the zeros of -4t^2+32t=-132

The zeros of -4t^2+32t=-132 are t=-3,t=11

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- 4 t^{2} + 32 t = - 132

t = - 3, t = 11

Solution steps

- 4 t^{2} + 32 t = - 132

Move

132

to the left side

- 4 t^{2} + 32 t + 132 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 32 \pm 3 2 ^{2} - 4 ( - 4 ) \cdot 132}{2 ( - 4 )}

3 2^{2} - 4 (- 4) \cdot 132 = 56

t_{1, 2} = \frac{- 32 \pm 56}{2 ( - 4 )}

Separate the solutions

t_{1} = \frac{- 32 + 56}{2 ( - 4 )}, t_{2} = \frac{- 32 - 56}{2 ( - 4 )}

t = \frac{- 32 + 56}{2 ( - 4 )} : - 3

t = \frac{- 32 - 56}{2 ( - 4 )} : 11

The solutions to the quadratic equation are:

t = - 3, t = 11

x^{2} + (4 x + 16)^{2} = (4 x + 17)^{2}

x^{2} + 2 x = 80

x^{2} + 2 x = 15

- 24 = - 3 x^{2} - 6 x

27 x^{2} + 5 = 48 x

What are the solutions to the equation -4t^2+32t=-132 ?
The solutions to the equation -4t^2+32t=-132 are t=-3,t=11
Find the zeros of -4t^2+32t=-132
The zeros of -4t^2+32t=-132 are t=-3,t=11