What are the solutions to the equation 4.9t^2-5.2t-125=0 ?

The solutions to the equation 4.9t^2-5.2t-125=0 are t=(26+sqrt(61926))/(49),t=(26-sqrt(61926))/(49)

Find the zeros of 4.9t^2-5.2t-125=0

The zeros of 4.9t^2-5.2t-125=0 are t=(26+sqrt(61926))/(49),t=(26-sqrt(61926))/(49)

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4.9 t^{2} - 5.2 t - 125 = 0

t = \frac{26 + 61926}{49}, t = \frac{26 - 61926}{49}

Decimal Notation

t = 5.60917 \dots, t = - 4.54794 \dots

Solution steps

4.9 t^{2} - 5.2 t - 125 = 0

Multiply both sides by

10

49 t^{2} - 52 t - 1250 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 52 ) \pm ( - 52 ) ^{2} - 4 \cdot 49 ( - 1250 )}{2 \cdot 49}

(- 52)^{2} - 4 \cdot 49 (- 1250) = 261926

t_{1, 2} = \frac{- ( - 52 ) \pm 2 61926}{2 \cdot 49}

Separate the solutions

t_{1} = \frac{- ( - 52 ) + 2 61926}{2 \cdot 49}, t_{2} = \frac{- ( - 52 ) - 2 61926}{2 \cdot 49}

t = \frac{- ( - 52 ) + 2 61926}{2 \cdot 49} : \frac{26 + 61926}{49}

t = \frac{- ( - 52 ) - 2 61926}{2 \cdot 49} : \frac{26 - 61926}{49}

The solutions to the quadratic equation are:

t = \frac{26 + 61926}{49}, t = \frac{26 - 61926}{49}

6 x^{2} + 8 x - 8 = 0

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y, x^{2} - y^{2} = 0

4 x^{2} + 10 x = 36

y^{2} - 14 y + 42 = 0

What are the solutions to the equation 4.9t^2-5.2t-125=0 ?
The solutions to the equation 4.9t^2-5.2t-125=0 are t=(26+sqrt(61926))/(49),t=(26-sqrt(61926))/(49)
Find the zeros of 4.9t^2-5.2t-125=0
The zeros of 4.9t^2-5.2t-125=0 are t=(26+sqrt(61926))/(49),t=(26-sqrt(61926))/(49)