What are the solutions to the equation 0=40+5x-4.9x^2 ?

The solutions to the equation 0=40+5x-4.9x^2 are x=-(5(sqrt(809)-5))/(49),x=(5(5+sqrt(809)))/(49)

Find the zeros of 0=40+5x-4.9x^2

The zeros of 0=40+5x-4.9x^2 are x=-(5(sqrt(809)-5))/(49),x=(5(5+sqrt(809)))/(49)

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0 = 40 + 5 x - 4.9 x^{2}

x = - \frac{5 ( 809 - 5 )}{49}, x = \frac{5 ( 5 + 809 )}{49}

Decimal Notation

x = - 2.39213 \dots, x = 3.41254 \dots

Solution steps

0 = 40 + 5 x - 4.9 x^{2}

Multiply both sides by

10

0 = 400 + 50 x - 49 x^{2}

Switch sides

400 + 50 x - 49 x^{2} = 0

Write in the standard form

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

- 49 x^{2} + 50 x + 400 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 50 \pm 5 0 ^{2} - 4 ( - 49 ) \cdot 400}{2 ( - 49 )}

5 0^{2} - 4 (- 49) \cdot 400 = 10809

x_{1, 2} = \frac{- 50 \pm 10 809}{2 ( - 49 )}

Separate the solutions

x_{1} = \frac{- 50 + 10 809}{2 ( - 49 )}, x_{2} = \frac{- 50 - 10 809}{2 ( - 49 )}

x = \frac{- 50 + 10 809}{2 ( - 49 )} : - \frac{5 ( 809 - 5 )}{49}

x = \frac{- 50 - 10 809}{2 ( - 49 )} : \frac{5 ( 5 + 809 )}{49}

The solutions to the quadratic equation are:

x = - \frac{5 ( 809 - 5 )}{49}, x = \frac{5 ( 5 + 809 )}{49}

x, x^{2} + y^{2} = a

2 x (x - 2) = 6 x^{2} - 5 x

2 w^{2} - 5 w = 33

0.8 x^{2} - 24 x + 200 = 0

x^{2} + 12 x + 144 = 0

What are the solutions to the equation 0=40+5x-4.9x^2 ?
The solutions to the equation 0=40+5x-4.9x^2 are x=-(5(sqrt(809)-5))/(49),x=(5(5+sqrt(809)))/(49)
Find the zeros of 0=40+5x-4.9x^2
The zeros of 0=40+5x-4.9x^2 are x=-(5(sqrt(809)-5))/(49),x=(5(5+sqrt(809)))/(49)