What are the solutions to the equation 1990(6)=4x^2+5x+2 ?

The solutions to the equation 1990(6)=4x^2+5x+2 are x=(-5+sqrt(191033))/8 ,x=(-5-sqrt(191033))/8

Find the zeros of 1990(6)=4x^2+5x+2

The zeros of 1990(6)=4x^2+5x+2 are x=(-5+sqrt(191033))/8 ,x=(-5-sqrt(191033))/8

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1990 (6) = 4 x^{2} + 5 x + 2

x = \frac{- 5 + 191033}{8}, x = \frac{- 5 - 191033}{8}

Solution steps

1990 (6) = 4 x^{2} + 5 x + 2

Expand

1990 (6) : 11940

11940 = 4 x^{2} + 5 x + 2

Switch sides

4 x^{2} + 5 x + 2 = 11940

Move

11940

to the left side

4 x^{2} + 5 x - 11938 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 4 ( - 11938 )}{2 \cdot 4}

5^{2} - 4 \cdot 4 (- 11938) = 191033

x_{1, 2} = \frac{- 5 \pm 191033}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 5 + 191033}{2 \cdot 4}, x_{2} = \frac{- 5 - 191033}{2 \cdot 4}

x = \frac{- 5 + 191033}{2 \cdot 4} : \frac{- 5 + 191033}{8}

x = \frac{- 5 - 191033}{2 \cdot 4} : \frac{- 5 - 191033}{8}

The solutions to the quadratic equation are:

x = \frac{- 5 + 191033}{8}, x = \frac{- 5 - 191033}{8}

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\frac{1 ^{6} + C}{e ^{(2 l n (1))}} = 1

643 x + 4600 x^{2} - 1 = 0

x^{3} - 5 x^{2} - 18 x + 72

What are the solutions to the equation 1990(6)=4x^2+5x+2 ?
The solutions to the equation 1990(6)=4x^2+5x+2 are x=(-5+sqrt(191033))/8 ,x=(-5-sqrt(191033))/8
Find the zeros of 1990(6)=4x^2+5x+2
The zeros of 1990(6)=4x^2+5x+2 are x=(-5+sqrt(191033))/8 ,x=(-5-sqrt(191033))/8