What are the solutions to the equation -2t(t+3)=-7 ?

The solutions to the equation -2t(t+3)=-7 are t=-(3+sqrt(23))/2 ,t=(sqrt(23)-3)/2

Find the zeros of -2t(t+3)=-7

The zeros of -2t(t+3)=-7 are t=-(3+sqrt(23))/2 ,t=(sqrt(23)-3)/2

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- 2 t (t + 3) = - 7

t = - \frac{3 + 23}{2}, t = \frac{23 - 3}{2}

Decimal Notation

t = - 3.89791 \dots, t = 0.89791 \dots

Solution steps

- 2 t (t + 3) = - 7

Expand

- 2 t (t + 3) : - 2 t^{2} - 6 t

- 2 t^{2} - 6 t = - 7

Move

7

to the left side

- 2 t^{2} - 6 t + 7 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 ( - 2 ) \cdot 7}{2 ( - 2 )}

(- 6)^{2} - 4 (- 2) \cdot 7 = 223

t_{1, 2} = \frac{- ( - 6 ) \pm 2 23}{2 ( - 2 )}

Separate the solutions

t_{1} = \frac{- ( - 6 ) + 2 23}{2 ( - 2 )}, t_{2} = \frac{- ( - 6 ) - 2 23}{2 ( - 2 )}

t = \frac{- ( - 6 ) + 2 23}{2 ( - 2 )} : - \frac{3 + 23}{2}

t = \frac{- ( - 6 ) - 2 23}{2 ( - 2 )} : \frac{23 - 3}{2}

The solutions to the quadratic equation are:

t = - \frac{3 + 23}{2}, t = \frac{23 - 3}{2}

7 n^{2} + 84 n + 250 = 5

3 x^{2} = 225

3 x^{2} = 2 - x

x^{2} - x + \frac{5}{2} = 0

1 + x^{2} = 16

What are the solutions to the equation -2t(t+3)=-7 ?
The solutions to the equation -2t(t+3)=-7 are t=-(3+sqrt(23))/2 ,t=(sqrt(23)-3)/2
Find the zeros of -2t(t+3)=-7
The zeros of -2t(t+3)=-7 are t=-(3+sqrt(23))/2 ,t=(sqrt(23)-3)/2