What are the solutions to the equation 4t^2+12t-3=5t ?

The solutions to the equation 4t^2+12t-3=5t are t=(-7+sqrt(97))/8 ,t=(-7-sqrt(97))/8

Find the zeros of 4t^2+12t-3=5t

The zeros of 4t^2+12t-3=5t are t=(-7+sqrt(97))/8 ,t=(-7-sqrt(97))/8

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4 t^{2} + 12 t - 3 = 5 t

t = \frac{- 7 + 97}{8}, t = \frac{- 7 - 97}{8}

Decimal

t = 0.35610 \dots, t = - 2.10610 \dots

Solution steps

4 t^{2} + 12 t - 3 = 5 t

Move

5 t

to the left side

4 t^{2} + 7 t - 3 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 4 ( - 3 )}{2 \cdot 4}

7^{2} - 4 \cdot 4 (- 3) = 97

t_{1, 2} = \frac{- 7 \pm 97}{2 \cdot 4}

Separate the solutions

t_{1} = \frac{- 7 + 97}{2 \cdot 4}, t_{2} = \frac{- 7 - 97}{2 \cdot 4}

t = \frac{- 7 + 97}{2 \cdot 4} : \frac{- 7 + 97}{8}

t = \frac{- 7 - 97}{2 \cdot 4} : \frac{- 7 - 97}{8}

The solutions to the quadratic equation are:

t = \frac{- 7 + 97}{8}, t = \frac{- 7 - 97}{8}

x^{2} - 10 x = - 14

18 = y^{2}

2560 = 4000 b^{2}

x^{2} + 6 x + 3

x^{2} + 6 x + 5

What are the solutions to the equation 4t^2+12t-3=5t ?
The solutions to the equation 4t^2+12t-3=5t are t=(-7+sqrt(97))/8 ,t=(-7-sqrt(97))/8
Find the zeros of 4t^2+12t-3=5t
The zeros of 4t^2+12t-3=5t are t=(-7+sqrt(97))/8 ,t=(-7-sqrt(97))/8