What are the solutions to the equation 5t^2+2t-12=0 ?

The solutions to the equation 5t^2+2t-12=0 are t=(-1+sqrt(61))/5 ,t=-(1+sqrt(61))/5

Find the zeros of 5t^2+2t-12=0

The zeros of 5t^2+2t-12=0 are t=(-1+sqrt(61))/5 ,t=-(1+sqrt(61))/5

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5 t^{2} + 2 t - 12 = 0

t = \frac{- 1 + 61}{5}, t = - \frac{1 + 61}{5}

Decimal Notation

t = 1.36204 \dots, t = - 1.76204 \dots

Solution steps

5 t^{2} + 2 t - 12 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 5 ( - 12 )}{2 \cdot 5}

2^{2} - 4 \cdot 5 (- 12) = 261

t_{1, 2} = \frac{- 2 \pm 2 61}{2 \cdot 5}

Separate the solutions

t_{1} = \frac{- 2 + 2 61}{2 \cdot 5}, t_{2} = \frac{- 2 - 2 61}{2 \cdot 5}

t = \frac{- 2 + 2 61}{2 \cdot 5} : \frac{- 1 + 61}{5}

t = \frac{- 2 - 2 61}{2 \cdot 5} : - \frac{1 + 61}{5}

The solutions to the quadratic equation are:

t = \frac{- 1 + 61}{5}, t = - \frac{1 + 61}{5}

x^{2} - 19 x + 88

\frac{5 x ^{3} y ^{3} + 3 x ^{2} y ^{2}}{x ^{2} y ^{2}}

\frac{5}{5}

∣ 4 x - 3 ∣ = ∣ 2 x - 2 ∣

3 y + 2 = 11

What are the solutions to the equation 5t^2+2t-12=0 ?
The solutions to the equation 5t^2+2t-12=0 are t=(-1+sqrt(61))/5 ,t=-(1+sqrt(61))/5
Find the zeros of 5t^2+2t-12=0
The zeros of 5t^2+2t-12=0 are t=(-1+sqrt(61))/5 ,t=-(1+sqrt(61))/5