What are the solutions to the equation 4.9x^2-14x-18.5=0 ?

The solutions to the equation 4.9x^2-14x-18.5=0 are x=(10+sqrt(285))/7 ,x=(10-sqrt(285))/7

Find the zeros of 4.9x^2-14x-18.5=0

The zeros of 4.9x^2-14x-18.5=0 are x=(10+sqrt(285))/7 ,x=(10-sqrt(285))/7

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4.9 x^{2} - 14 x - 18.5 = 0

x = \frac{10 + 285}{7}, x = \frac{10 - 285}{7}

Decimal Notation

x = 3.84027 \dots, x = - 0.98313 \dots

Solution steps

4.9 x^{2} - 14 x - 18.5 = 0

Multiply both sides by

10

49 x^{2} - 140 x - 185 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 140 ) \pm ( - 140 ) ^{2} - 4 \cdot 49 ( - 185 )}{2 \cdot 49}

(- 140)^{2} - 4 \cdot 49 (- 185) = 14285

x_{1, 2} = \frac{- ( - 140 ) \pm 14 285}{2 \cdot 49}

Separate the solutions

x_{1} = \frac{- ( - 140 ) + 14 285}{2 \cdot 49}, x_{2} = \frac{- ( - 140 ) - 14 285}{2 \cdot 49}

x = \frac{- ( - 140 ) + 14 285}{2 \cdot 49} : \frac{10 + 285}{7}

x = \frac{- ( - 140 ) - 14 285}{2 \cdot 49} : \frac{10 - 285}{7}

The solutions to the quadratic equation are:

x = \frac{10 + 285}{7}, x = \frac{10 - 285}{7}

(x^{3} y^{2}) (\frac{a y - b x}{ab})

0.5 x^{2} + 4 = 0

\frac{\frac{1}{7}}{3}

a^{2} + 1. 5^{2} = 0. 5^{2}

7 \leq \frac{x + 2}{4}

What are the solutions to the equation 4.9x^2-14x-18.5=0 ?
The solutions to the equation 4.9x^2-14x-18.5=0 are x=(10+sqrt(285))/7 ,x=(10-sqrt(285))/7
Find the zeros of 4.9x^2-14x-18.5=0
The zeros of 4.9x^2-14x-18.5=0 are x=(10+sqrt(285))/7 ,x=(10-sqrt(285))/7