What are the solutions to the equation 49x^2-140x+100=0 ?

The solutions to the equation 49x^2-140x+100=0 are x= 10/7

Find the zeros of 49x^2-140x+100=0

The zeros of 49x^2-140x+100=0 are x= 10/7

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49 x^{2} - 140 x + 100 = 0

x = \frac{10}{7}

Decimal Notation

x = 1.42857 \dots

Solution steps

49 x^{2} - 140 x + 100 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 140 ) \pm ( - 140 ) ^{2} - 4 \cdot 49 \cdot 100}{2 \cdot 49}

(- 140)^{2} - 4 \cdot 49 \cdot 100 = 0

x_{1, 2} = \frac{- ( - 140 ) \pm 0}{2 \cdot 49}

x = \frac{- ( - 140 )}{2 \cdot 49}

\frac{- ( - 140 )}{2 \cdot 49} = \frac{10}{7}

x = \frac{10}{7}

The solution to the quadratic equation is:

x = \frac{10}{7}

(5 y + 2) (1 + y) = 0

0 = - 16 t^{2} + 80 t + 4

x, x^{2} y^{2} + 1 = 2 x y

29 (x + 1) + 3 = - x^{2} + 27 x + 6

\frac{350 x ^{2}}{2} = 6.64 \cdot 1257 \cdot (540 - x)

What are the solutions to the equation 49x^2-140x+100=0 ?
The solutions to the equation 49x^2-140x+100=0 are x= 10/7
Find the zeros of 49x^2-140x+100=0
The zeros of 49x^2-140x+100=0 are x= 10/7