What are the solutions to the equation 100x^2-140x=-28 ?

The solutions to the equation 100x^2-140x=-28 are x=(7+sqrt(21))/(10),x=(7-sqrt(21))/(10)

Find the zeros of 100x^2-140x=-28

The zeros of 100x^2-140x=-28 are x=(7+sqrt(21))/(10),x=(7-sqrt(21))/(10)

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100 x^{2} - 140 x = - 28

x = \frac{7 + 21}{10}, x = \frac{7 - 21}{10}

Decimal Notation

x = 1.15825 \dots, x = 0.24174 \dots

Solution steps

100 x^{2} - 140 x = - 28

Move

28

to the left side

100 x^{2} - 140 x + 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 140 ) \pm ( - 140 ) ^{2} - 4 \cdot 100 \cdot 28}{2 \cdot 100}

(- 140)^{2} - 4 \cdot 100 \cdot 28 = 2021

x_{1, 2} = \frac{- ( - 140 ) \pm 20 21}{2 \cdot 100}

Separate the solutions

x_{1} = \frac{- ( - 140 ) + 20 21}{2 \cdot 100}, x_{2} = \frac{- ( - 140 ) - 20 21}{2 \cdot 100}

x = \frac{- ( - 140 ) + 20 21}{2 \cdot 100} : \frac{7 + 21}{10}

x = \frac{- ( - 140 ) - 20 21}{2 \cdot 100} : \frac{7 - 21}{10}

The solutions to the quadratic equation are:

x = \frac{7 + 21}{10}, x = \frac{7 - 21}{10}

- 5 x^{2} + 2 x + 11 = 0

2 x^{2} - 2 x + 15 = 0

837.2 = 0.68 \cdot 2.14 \cdot 100 \cdot x \cdot (9.5 - 0.4 \cdot x)

12 x^{2} - 48 x = - 39

(t + 1)^{2} = - 9

What are the solutions to the equation 100x^2-140x=-28 ?
The solutions to the equation 100x^2-140x=-28 are x=(7+sqrt(21))/(10),x=(7-sqrt(21))/(10)
Find the zeros of 100x^2-140x=-28
The zeros of 100x^2-140x=-28 are x=(7+sqrt(21))/(10),x=(7-sqrt(21))/(10)