What are the solutions to the equation 4.9t^2+11t-18=0 ?

The solutions to the equation 4.9t^2+11t-18=0 are t=(-55+sqrt(11845))/(49),t=-(55+sqrt(11845))/(49)

Find the zeros of 4.9t^2+11t-18=0

The zeros of 4.9t^2+11t-18=0 are t=(-55+sqrt(11845))/(49),t=-(55+sqrt(11845))/(49)

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4.9 t^{2} + 11 t - 18 = 0

t = \frac{- 55 + 11845}{49}, t = - \frac{55 + 11845}{49}

Decimal Notation

t = 1.09866 \dots, t = - 3.34356 \dots

Solution steps

4.9 t^{2} + 11 t - 18 = 0

Multiply both sides by

10

49 t^{2} + 110 t - 180 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 110 \pm 11 0 ^{2} - 4 \cdot 49 ( - 180 )}{2 \cdot 49}

11 0^{2} - 4 \cdot 49 (- 180) = 211845

t_{1, 2} = \frac{- 110 \pm 2 11845}{2 \cdot 49}

Separate the solutions

t_{1} = \frac{- 110 + 2 11845}{2 \cdot 49}, t_{2} = \frac{- 110 - 2 11845}{2 \cdot 49}

t = \frac{- 110 + 2 11845}{2 \cdot 49} : \frac{- 55 + 11845}{49}

t = \frac{- 110 - 2 11845}{2 \cdot 49} : - \frac{55 + 11845}{49}

The solutions to the quadratic equation are:

t = \frac{- 55 + 11845}{49}, t = - \frac{55 + 11845}{49}

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What are the solutions to the equation 4.9t^2+11t-18=0 ?
The solutions to the equation 4.9t^2+11t-18=0 are t=(-55+sqrt(11845))/(49),t=-(55+sqrt(11845))/(49)
Find the zeros of 4.9t^2+11t-18=0
The zeros of 4.9t^2+11t-18=0 are t=(-55+sqrt(11845))/(49),t=-(55+sqrt(11845))/(49)