What are the solutions to the equation 6-2t^2=9t+15 ?

The solutions to the equation 6-2t^2=9t+15 are t=-3,t=-3/2

Find the zeros of 6-2t^2=9t+15

The zeros of 6-2t^2=9t+15 are t=-3,t=-3/2

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6 - 2 t^{2} = 9 t + 15

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

Decimal Notation

t = - 3, t = - 1.5

Solution steps

6 - 2 t^{2} = 9 t + 15

Move

15

to the left side

- 2 t^{2} - 9 = 9 t

Move

9 t

to the left side

- 2 t^{2} - 9 - 9 t = 0

Write in the standard form

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

- 2 t^{2} - 9 t - 9 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 ( - 2 ) ( - 9 )}{2 ( - 2 )}

(- 9)^{2} - 4 (- 2) (- 9) = 3

t_{1, 2} = \frac{- ( - 9 ) \pm 3}{2 ( - 2 )}

Separate the solutions

t_{1} = \frac{- ( - 9 ) + 3}{2 ( - 2 )}, t_{2} = \frac{- ( - 9 ) - 3}{2 ( - 2 )}

t = \frac{- ( - 9 ) + 3}{2 ( - 2 )} : - 3

t = \frac{- ( - 9 ) - 3}{2 ( - 2 )} : - \frac{3}{2}

The solutions to the quadratic equation are:

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

π r^{2} = 2305

(x - y) (x - y)

3 x - 8 \geq - 14

(\frac{2 a ^{- 1} b}{a ^{2} b ^{- 3}})^{- 2}

- \frac{2}{3}

What are the solutions to the equation 6-2t^2=9t+15 ?
The solutions to the equation 6-2t^2=9t+15 are t=-3,t=-3/2
Find the zeros of 6-2t^2=9t+15
The zeros of 6-2t^2=9t+15 are t=-3,t=-3/2