What are the solutions to the equation 0=18+6t-4t^2 ?

The solutions to the equation 0=18+6t-4t^2 are t=-3/2 ,t=3

Find the zeros of 0=18+6t-4t^2

The zeros of 0=18+6t-4t^2 are t=-3/2 ,t=3

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0 = 18 + 6 t - 4 t^{2}

t = - \frac{3}{2}, t = 3

Decimal Notation

t = - 1.5, t = 3

Solution steps

0 = 18 + 6 t - 4 t^{2}

Switch sides

18 + 6 t - 4 t^{2} = 0

Write in the standard form

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

- 4 t^{2} + 6 t + 18 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 ( - 4 ) \cdot 18}{2 ( - 4 )}

6^{2} - 4 (- 4) \cdot 18 = 18

t_{1, 2} = \frac{- 6 \pm 18}{2 ( - 4 )}

Separate the solutions

t_{1} = \frac{- 6 + 18}{2 ( - 4 )}, t_{2} = \frac{- 6 - 18}{2 ( - 4 )}

t = \frac{- 6 + 18}{2 ( - 4 )} : - \frac{3}{2}

t = \frac{- 6 - 18}{2 ( - 4 )} : 3

The solutions to the quadratic equation are:

t = - \frac{3}{2}, t = 3

\frac{1}{4} x^{2} = - \frac{1}{2} x + 2

4 = x^{2} + 3

t^{2} + 3 = 0

x^{2} = 11 x + 126

s^{2} + 16 s + 100 = 0

What are the solutions to the equation 0=18+6t-4t^2 ?
The solutions to the equation 0=18+6t-4t^2 are t=-3/2 ,t=3
Find the zeros of 0=18+6t-4t^2
The zeros of 0=18+6t-4t^2 are t=-3/2 ,t=3