What are the solutions to the equation 0=-16t^2-16t+32 ?

The solutions to the equation 0=-16t^2-16t+32 are t=-2,t=1

Find the zeros of 0=-16t^2-16t+32

The zeros of 0=-16t^2-16t+32 are t=-2,t=1

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0 = - 16 t^{2} - 16 t + 32

t = - 2, t = 1

Solution steps

0 = - 16 t^{2} - 16 t + 32

Switch sides

- 16 t^{2} - 16 t + 32 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 ( - 16 ) \cdot 32}{2 ( - 16 )}

(- 16)^{2} - 4 (- 16) \cdot 32 = 48

t_{1, 2} = \frac{- ( - 16 ) \pm 48}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- ( - 16 ) + 48}{2 ( - 16 )}, t_{2} = \frac{- ( - 16 ) - 48}{2 ( - 16 )}

t = \frac{- ( - 16 ) + 48}{2 ( - 16 )} : - 2

t = \frac{- ( - 16 ) - 48}{2 ( - 16 )} : 1

The solutions to the quadratic equation are:

t = - 2, t = 1

∣ t + 15 ∣ = 2 t + 25

r^{2} + r - 1 = 0

2 (9)^{\frac{1}{2}}

r^{2} + 200 r + 10 = 0

(a - b) \cdot z + 3 \cdot (b - a)

What are the solutions to the equation 0=-16t^2-16t+32 ?
The solutions to the equation 0=-16t^2-16t+32 are t=-2,t=1
Find the zeros of 0=-16t^2-16t+32
The zeros of 0=-16t^2-16t+32 are t=-2,t=1