What are the solutions to the equation 20-2x^2-5x=4 ?

The solutions to the equation 20-2x^2-5x=4 are x=-(5+3sqrt(17))/4 ,x=(3sqrt(17)-5)/4

Find the zeros of 20-2x^2-5x=4

The zeros of 20-2x^2-5x=4 are x=-(5+3sqrt(17))/4 ,x=(3sqrt(17)-5)/4

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20 - 2 x^{2} - 5 x = 4

x = - \frac{5 + 3 17}{4}, x = \frac{3 17 - 5}{4}

Decimal Notation

x = - 4.34232 \dots, x = 1.84232 \dots

Solution steps

20 - 2 x^{2} - 5 x = 4

Move

4

to the left side

- 2 x^{2} - 5 x + 16 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( - 2 ) \cdot 16}{2 ( - 2 )}

(- 5)^{2} - 4 (- 2) \cdot 16 = 317

x_{1, 2} = \frac{- ( - 5 ) \pm 3 17}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 3 17}{2 ( - 2 )}, x_{2} = \frac{- ( - 5 ) - 3 17}{2 ( - 2 )}

x = \frac{- ( - 5 ) + 3 17}{2 ( - 2 )} : - \frac{5 + 3 17}{4}

x = \frac{- ( - 5 ) - 3 17}{2 ( - 2 )} : \frac{3 17 - 5}{4}

The solutions to the quadratic equation are:

x = - \frac{5 + 3 17}{4}, x = \frac{3 17 - 5}{4}

x^{2} - 20 x + 80 = 0

(m^{2} + 2 m + 2) = 0

(x - 3)^{2} + 1 = 2 x - 5

\frac{1}{4} (2 A) + 2 A x + B + A x^{2} + B x + C = x^{2} - 2 x

0.48 x^{2} - 1.88 x - 1.68 = 0

What are the solutions to the equation 20-2x^2-5x=4 ?
The solutions to the equation 20-2x^2-5x=4 are x=-(5+3sqrt(17))/4 ,x=(3sqrt(17)-5)/4
Find the zeros of 20-2x^2-5x=4
The zeros of 20-2x^2-5x=4 are x=-(5+3sqrt(17))/4 ,x=(3sqrt(17)-5)/4