What are the solutions to the equation 0.0625x^2-2x+10=0 ?

The solutions to the equation 0.0625x^2-2x+10=0 are x=(2+sqrt(1.5))/(0.125),x=(2-sqrt(1.5))/(0.125)

Find the zeros of 0.0625x^2-2x+10=0

The zeros of 0.0625x^2-2x+10=0 are x=(2+sqrt(1.5))/(0.125),x=(2-sqrt(1.5))/(0.125)

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0.0625 x^{2} - 2 x + 10 = 0

x = \frac{2 + 1.5}{0.125}, x = \frac{2 - 1.5}{0.125}

Decimal Notation

x = 25.79795 \dots, x = 6.20204 \dots

Solution steps

0.0625 x^{2} - 2 x + 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 0.0625 \cdot 10}{2 \cdot 0.0625}

(- 2)^{2} - 4 \cdot 0.0625 \cdot 10 = 1.5

x_{1, 2} = \frac{- ( - 2 ) \pm 1.5}{2 \cdot 0.0625}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 1.5}{2 \cdot 0.0625}, x_{2} = \frac{- ( - 2 ) - 1.5}{2 \cdot 0.0625}

x = \frac{- ( - 2 ) + 1.5}{2 \cdot 0.0625} : \frac{2 + 1.5}{0.125}

x = \frac{- ( - 2 ) - 1.5}{2 \cdot 0.0625} : \frac{2 - 1.5}{0.125}

The solutions to the quadratic equation are:

x = \frac{2 + 1.5}{0.125}, x = \frac{2 - 1.5}{0.125}

\frac{x ^{2} + 1}{5 x} \leq \frac{1}{2}

x^{3} + 4 x^{2} + 3 x + 12

(x - 2) (x + 1) (2 x + 3) \leq 0

\frac{5}{3 + 4 3}

∣ x + 1 ∣ + ∣ x - 1 ∣ = 2

What are the solutions to the equation 0.0625x^2-2x+10=0 ?
The solutions to the equation 0.0625x^2-2x+10=0 are x=(2+sqrt(1.5))/(0.125),x=(2-sqrt(1.5))/(0.125)
Find the zeros of 0.0625x^2-2x+10=0
The zeros of 0.0625x^2-2x+10=0 are x=(2+sqrt(1.5))/(0.125),x=(2-sqrt(1.5))/(0.125)