What are the solutions to the equation 12t-4.9t^2+70=0 ?

The solutions to the equation 12t-4.9t^2+70=0 are t=-(10(sqrt(379)-6))/(49),t=(10(6+sqrt(379)))/(49)

Find the zeros of 12t-4.9t^2+70=0

The zeros of 12t-4.9t^2+70=0 are t=-(10(sqrt(379)-6))/(49),t=(10(6+sqrt(379)))/(49)

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12 t - 4.9 t^{2} + 70 = 0

t = - \frac{10 ( 379 - 6 )}{49}, t = \frac{10 ( 6 + 379 )}{49}

Decimal Notation

t = - 2.74855 \dots, t = 5.19753 \dots

Solution steps

12 t - 4.9 t^{2} + 70 = 0

Multiply both sides by

10

120 t - 49 t^{2} + 700 = 0

Write in the standard form

a x^{2} + b x + c = 0

- 49 t^{2} + 120 t + 700 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 120 \pm 12 0 ^{2} - 4 ( - 49 ) \cdot 700}{2 ( - 49 )}

12 0^{2} - 4 (- 49) \cdot 700 = 20379

t_{1, 2} = \frac{- 120 \pm 20 379}{2 ( - 49 )}

Separate the solutions

t_{1} = \frac{- 120 + 20 379}{2 ( - 49 )}, t_{2} = \frac{- 120 - 20 379}{2 ( - 49 )}

t = \frac{- 120 + 20 379}{2 ( - 49 )} : - \frac{10 ( 379 - 6 )}{49}

t = \frac{- 120 - 20 379}{2 ( - 49 )} : \frac{10 ( 6 + 379 )}{49}

The solutions to the quadratic equation are:

t = - \frac{10 ( 379 - 6 )}{49}, t = \frac{10 ( 6 + 379 )}{49}

a^{2} = e^{2}

y, 9 x^{2} + 16 y^{2} = 49

- 24 x^{2} - 24 x + 288 = 0

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

(2 x + 5) (- m x + 9) = 0

What are the solutions to the equation 12t-4.9t^2+70=0 ?
The solutions to the equation 12t-4.9t^2+70=0 are t=-(10(sqrt(379)-6))/(49),t=(10(6+sqrt(379)))/(49)
Find the zeros of 12t-4.9t^2+70=0
The zeros of 12t-4.9t^2+70=0 are t=-(10(sqrt(379)-6))/(49),t=(10(6+sqrt(379)))/(49)