What are the solutions to the equation 4x^2-41x=73 ?

The solutions to the equation 4x^2-41x=73 are x=(41+sqrt(2849))/8 ,x=(41-sqrt(2849))/8

Find the zeros of 4x^2-41x=73

The zeros of 4x^2-41x=73 are x=(41+sqrt(2849))/8 ,x=(41-sqrt(2849))/8

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4 x^{2} - 41 x = 73

x = \frac{41 + 2849}{8}, x = \frac{41 - 2849}{8}

Decimal Notation

x = 11.79700 \dots, x = - 1.54700 \dots

Solution steps

4 x^{2} - 41 x = 73

Move

73

to the left side

4 x^{2} - 41 x - 73 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 41 ) \pm ( - 41 ) ^{2} - 4 \cdot 4 ( - 73 )}{2 \cdot 4}

(- 41)^{2} - 4 \cdot 4 (- 73) = 2849

x_{1, 2} = \frac{- ( - 41 ) \pm 2849}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 41 ) + 2849}{2 \cdot 4}, x_{2} = \frac{- ( - 41 ) - 2849}{2 \cdot 4}

x = \frac{- ( - 41 ) + 2849}{2 \cdot 4} : \frac{41 + 2849}{8}

x = \frac{- ( - 41 ) - 2849}{2 \cdot 4} : \frac{41 - 2849}{8}

The solutions to the quadratic equation are:

x = \frac{41 + 2849}{8}, x = \frac{41 - 2849}{8}

0 = 3 x^{2} - 5 x - 5

(x - 512) (x + 512) = 0

c^{2} - 17 c + 16 = - 2 c^{2}

2 v^{2} + 3 v - 5 = - 1

2 w (w - 1) = 5

What are the solutions to the equation 4x^2-41x=73 ?
The solutions to the equation 4x^2-41x=73 are x=(41+sqrt(2849))/8 ,x=(41-sqrt(2849))/8
Find the zeros of 4x^2-41x=73
The zeros of 4x^2-41x=73 are x=(41+sqrt(2849))/8 ,x=(41-sqrt(2849))/8