What are the solutions to the equation -4.9x^2+10x+2=0 ?

The solutions to the equation -4.9x^2+10x+2=0 are x=-(2(sqrt(870)-25))/(49),x=(2(25+sqrt(870)))/(49)

Find the zeros of -4.9x^2+10x+2=0

The zeros of -4.9x^2+10x+2=0 are x=-(2(sqrt(870)-25))/(49),x=(2(25+sqrt(870)))/(49)

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- 4.9 x^{2} + 10 x + 2 = 0

x = - \frac{2 ( 870 - 25 )}{49}, x = \frac{2 ( 25 + 870 )}{49}

Decimal Notation

x = - 0.18350 \dots, x = 2.22431 \dots

Solution steps

- 4.9 x^{2} + 10 x + 2 = 0

Multiply both sides by

10

- 49 x^{2} + 100 x + 20 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 100 \pm 10 0 ^{2} - 4 ( - 49 ) \cdot 20}{2 ( - 49 )}

10 0^{2} - 4 (- 49) \cdot 20 = 4870

x_{1, 2} = \frac{- 100 \pm 4 870}{2 ( - 49 )}

Separate the solutions

x_{1} = \frac{- 100 + 4 870}{2 ( - 49 )}, x_{2} = \frac{- 100 - 4 870}{2 ( - 49 )}

x = \frac{- 100 + 4 870}{2 ( - 49 )} : - \frac{2 ( 870 - 25 )}{49}

x = \frac{- 100 - 4 870}{2 ( - 49 )} : \frac{2 ( 25 + 870 )}{49}

The solutions to the quadratic equation are:

x = - \frac{2 ( 870 - 25 )}{49}, x = \frac{2 ( 25 + 870 )}{49}

5 y^{2} + 1 = 8

3 (3 x - 2) = (x - 4) (x + 4)

4 B = (t + 2)^{2}

6 a^{2} - 2 = 0

x, g = a (x - h)^{2} + 3

What are the solutions to the equation -4.9x^2+10x+2=0 ?
The solutions to the equation -4.9x^2+10x+2=0 are x=-(2(sqrt(870)-25))/(49),x=(2(25+sqrt(870)))/(49)
Find the zeros of -4.9x^2+10x+2=0
The zeros of -4.9x^2+10x+2=0 are x=-(2(sqrt(870)-25))/(49),x=(2(25+sqrt(870)))/(49)