What are the solutions to the equation 5t^2+7t-3=4t ?

The solutions to the equation 5t^2+7t-3=4t are t=(-3+sqrt(69))/(10),t=(-3-sqrt(69))/(10)

Find the zeros of 5t^2+7t-3=4t

The zeros of 5t^2+7t-3=4t are t=(-3+sqrt(69))/(10),t=(-3-sqrt(69))/(10)

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

t = \frac{- 3 + 69}{10}, t = \frac{- 3 - 69}{10}

Decimal Notation

t = 0.53066 \dots, t = - 1.13066 \dots

Solution steps

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

Move

4 t

to the left side

5 t^{2} + 3 t - 3 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 5 ( - 3 )}{2 \cdot 5}

3^{2} - 4 \cdot 5 (- 3) = 69

t_{1, 2} = \frac{- 3 \pm 69}{2 \cdot 5}

Separate the solutions

t_{1} = \frac{- 3 + 69}{2 \cdot 5}, t_{2} = \frac{- 3 - 69}{2 \cdot 5}

t = \frac{- 3 + 69}{2 \cdot 5} : \frac{- 3 + 69}{10}

t = \frac{- 3 - 69}{2 \cdot 5} : \frac{- 3 - 69}{10}

The solutions to the quadratic equation are:

t = \frac{- 3 + 69}{10}, t = \frac{- 3 - 69}{10}

x^{2} + (\frac{9}{8})^{2} - 1 = 0

- x^{2} + 5 x + 30 = 0

4 v^{2} + 4 v + 1 = 0

(x - 5) (x + 1) + 5 = 0

x^{2} = 1 0^{2} + 1 0^{2}

What are the solutions to the equation 5t^2+7t-3=4t ?
The solutions to the equation 5t^2+7t-3=4t are t=(-3+sqrt(69))/(10),t=(-3-sqrt(69))/(10)
Find the zeros of 5t^2+7t-3=4t
The zeros of 5t^2+7t-3=4t are t=(-3+sqrt(69))/(10),t=(-3-sqrt(69))/(10)