What are the solutions to the equation 3.1x+6.2=1.05x^2 ?

The solutions to the equation 3.1x+6.2=1.05x^2 are x=(31+sqrt(3565))/(21),x=(31-sqrt(3565))/(21)

Find the zeros of 3.1x+6.2=1.05x^2

The zeros of 3.1x+6.2=1.05x^2 are x=(31+sqrt(3565))/(21),x=(31-sqrt(3565))/(21)

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3.1 x + 6.2 = 1.05 x^{2}

x = \frac{31 + 3565}{21}, x = \frac{31 - 3565}{21}

Decimal

x = 4.31941 \dots, x = - 1.36702 \dots

Solution steps

3.1 x + 6.2 = 1.05 x^{2}

Multiply both sides by

100

310 x + 620 = 105 x^{2}

Switch sides

105 x^{2} = 310 x + 620

Move

620

to the left side

105 x^{2} - 620 = 310 x

Move

310 x

to the left side

105 x^{2} - 620 - 310 x = 0

Write in the standard form

a x^{2} + b x + c = 0

105 x^{2} - 310 x - 620 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 310 ) \pm ( - 310 ) ^{2} - 4 \cdot 105 ( - 620 )}{2 \cdot 105}

(- 310)^{2} - 4 \cdot 105 (- 620) = 103565

x_{1, 2} = \frac{- ( - 310 ) \pm 10 3565}{2 \cdot 105}

Separate the solutions

x_{1} = \frac{- ( - 310 ) + 10 3565}{2 \cdot 105}, x_{2} = \frac{- ( - 310 ) - 10 3565}{2 \cdot 105}

x = \frac{- ( - 310 ) + 10 3565}{2 \cdot 105} : \frac{31 + 3565}{21}

x = \frac{- ( - 310 ) - 10 3565}{2 \cdot 105} : \frac{31 - 3565}{21}

The solutions to the quadratic equation are:

x = \frac{31 + 3565}{21}, x = \frac{31 - 3565}{21}

(2 x - 1) (3 x + 5) = (x + 2) (2 x - 1)

(256 x^{16})^{\frac{1}{4}}

\frac{1}{- \frac{5}{13}}

- 11 \cdot - 7

\frac{1}{3} (6)^{3}

What are the solutions to the equation 3.1x+6.2=1.05x^2 ?
The solutions to the equation 3.1x+6.2=1.05x^2 are x=(31+sqrt(3565))/(21),x=(31-sqrt(3565))/(21)
Find the zeros of 3.1x+6.2=1.05x^2
The zeros of 3.1x+6.2=1.05x^2 are x=(31+sqrt(3565))/(21),x=(31-sqrt(3565))/(21)