What are the solutions to the equation 2t^2+4t-6=-3t ?

The solutions to the equation 2t^2+4t-6=-3t are t=(-7+sqrt(97))/4 ,t=(-7-sqrt(97))/4

Find the zeros of 2t^2+4t-6=-3t

The zeros of 2t^2+4t-6=-3t are t=(-7+sqrt(97))/4 ,t=(-7-sqrt(97))/4

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2 t^{2} + 4 t - 6 = - 3 t

t = \frac{- 7 + 97}{4}, t = \frac{- 7 - 97}{4}

Decimal Notation

t = 0.71221 \dots, t = - 4.21221 \dots

Solution steps

2 t^{2} + 4 t - 6 = - 3 t

Move

3 t

to the left side

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

Solve with the quadratic formula

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

7^{2} - 4 \cdot 2 (- 6) = 97

t_{1, 2} = \frac{- 7 \pm 97}{2 \cdot 2}

Separate the solutions

t_{1} = \frac{- 7 + 97}{2 \cdot 2}, t_{2} = \frac{- 7 - 97}{2 \cdot 2}

t = \frac{- 7 + 97}{2 \cdot 2} : \frac{- 7 + 97}{4}

t = \frac{- 7 - 97}{2 \cdot 2} : \frac{- 7 - 97}{4}

The solutions to the quadratic equation are:

t = \frac{- 7 + 97}{4}, t = \frac{- 7 - 97}{4}

4 x - 2 x^{2} = 1

0.57 x^{2} + 0.41 x - 0.87 = 0

a^{3} - (a - 1)^{3} = 169

2 a^{2} - \frac{7}{2} a - 18 = 0

2 t^{2} + 11 t + 5 = 0

What are the solutions to the equation 2t^2+4t-6=-3t ?
The solutions to the equation 2t^2+4t-6=-3t are t=(-7+sqrt(97))/4 ,t=(-7-sqrt(97))/4
Find the zeros of 2t^2+4t-6=-3t
The zeros of 2t^2+4t-6=-3t are t=(-7+sqrt(97))/4 ,t=(-7-sqrt(97))/4