What are the solutions to the equation 20x^2+12x+10=50 ?

The solutions to the equation 20x^2+12x+10=50 are x=(-3+sqrt(209))/(10),x=-(3+sqrt(209))/(10)

Find the zeros of 20x^2+12x+10=50

The zeros of 20x^2+12x+10=50 are x=(-3+sqrt(209))/(10),x=-(3+sqrt(209))/(10)

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20 x^{2} + 12 x + 10 = 50

x = \frac{- 3 + 209}{10}, x = - \frac{3 + 209}{10}

Decimal Notation

x = 1.14568 \dots, x = - 1.74568 \dots

Solution steps

20 x^{2} + 12 x + 10 = 50

Move

50

to the left side

20 x^{2} + 12 x - 40 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 20 ( - 40 )}{2 \cdot 20}

1 2^{2} - 4 \cdot 20 (- 40) = 4209

x_{1, 2} = \frac{- 12 \pm 4 209}{2 \cdot 20}

Separate the solutions

x_{1} = \frac{- 12 + 4 209}{2 \cdot 20}, x_{2} = \frac{- 12 - 4 209}{2 \cdot 20}

x = \frac{- 12 + 4 209}{2 \cdot 20} : \frac{- 3 + 209}{10}

x = \frac{- 12 - 4 209}{2 \cdot 20} : - \frac{3 + 209}{10}

The solutions to the quadratic equation are:

x = \frac{- 3 + 209}{10}, x = - \frac{3 + 209}{10}

4 = \frac{u}{10}

\frac{4}{6} \cdot \frac{3}{5}

\frac{1}{2} (10 - n) < 2 n

2^{5} \cdot 2^{2}

8 y^{2} + 6 y^{3} - 2 y^{4}

What are the solutions to the equation 20x^2+12x+10=50 ?
The solutions to the equation 20x^2+12x+10=50 are x=(-3+sqrt(209))/(10),x=-(3+sqrt(209))/(10)
Find the zeros of 20x^2+12x+10=50
The zeros of 20x^2+12x+10=50 are x=(-3+sqrt(209))/(10),x=-(3+sqrt(209))/(10)