What are the solutions to the equation 4t^2+17t+18=t^2 ?

The solutions to the equation 4t^2+17t+18=t^2 are t=(-17+sqrt(73))/6 ,t=(-17-sqrt(73))/6

Find the zeros of 4t^2+17t+18=t^2

The zeros of 4t^2+17t+18=t^2 are t=(-17+sqrt(73))/6 ,t=(-17-sqrt(73))/6

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4 t^{2} + 17 t + 18 = t^{2}

t = \frac{- 17 + 73}{6}, t = \frac{- 17 - 73}{6}

Solution steps

4 t^{2} + 17 t + 18 = t^{2}

Move

t^{2}

to the left side

3 t^{2} + 17 t + 18 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 17 \pm 1 7 ^{2} - 4 \cdot 3 \cdot 18}{2 \cdot 3}

1 7^{2} - 4 \cdot 3 \cdot 18 = 73

t_{1, 2} = \frac{- 17 \pm 73}{2 \cdot 3}

Separate the solutions

t_{1} = \frac{- 17 + 73}{2 \cdot 3}, t_{2} = \frac{- 17 - 73}{2 \cdot 3}

t = \frac{- 17 + 73}{2 \cdot 3} : \frac{- 17 + 73}{6}

t = \frac{- 17 - 73}{2 \cdot 3} : \frac{- 17 - 73}{6}

The solutions to the quadratic equation are:

t = \frac{- 17 + 73}{6}, t = \frac{- 17 - 73}{6}

5 a^{2} - 15 a + 11 = 0

3 x^{2} - 5 = 20 - x^{2}

\frac{y ^{2}}{2} - 3 y = 8

2 p^{2} + 13 p + 17 = 0

x^{2} + 16 x = - 59

What are the solutions to the equation 4t^2+17t+18=t^2 ?
The solutions to the equation 4t^2+17t+18=t^2 are t=(-17+sqrt(73))/6 ,t=(-17-sqrt(73))/6
Find the zeros of 4t^2+17t+18=t^2
The zeros of 4t^2+17t+18=t^2 are t=(-17+sqrt(73))/6 ,t=(-17-sqrt(73))/6