What are the solutions to the equation -4x^2-20x+16=0 ?

The solutions to the equation -4x^2-20x+16=0 are x=-(5+sqrt(41))/2 ,x=(sqrt(41)-5)/2

Find the zeros of -4x^2-20x+16=0

The zeros of -4x^2-20x+16=0 are x=-(5+sqrt(41))/2 ,x=(sqrt(41)-5)/2

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- 4 x^{2} - 20 x + 16 = 0

x = - \frac{5 + 41}{2}, x = \frac{41 - 5}{2}

Decimal Notation

x = - 5.70156 \dots, x = 0.70156 \dots

Solution steps

- 4 x^{2} - 20 x + 16 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 ( - 4 ) \cdot 16}{2 ( - 4 )}

(- 20)^{2} - 4 (- 4) \cdot 16 = 441

x_{1, 2} = \frac{- ( - 20 ) \pm 4 41}{2 ( - 4 )}

Separate the solutions

x_{1} = \frac{- ( - 20 ) + 4 41}{2 ( - 4 )}, x_{2} = \frac{- ( - 20 ) - 4 41}{2 ( - 4 )}

x = \frac{- ( - 20 ) + 4 41}{2 ( - 4 )} : - \frac{5 + 41}{2}

x = \frac{- ( - 20 ) - 4 41}{2 ( - 4 )} : \frac{41 - 5}{2}

The solutions to the quadratic equation are:

x = - \frac{5 + 41}{2}, x = \frac{41 - 5}{2}

10 - 3 x - x^{2} = 0

2 x (5 x + 5) = 88

a^{2} + 7 = 43

4 a^{2} - a - 8 = 0

5 x^{2} + 17 x + 4 = 0

What are the solutions to the equation -4x^2-20x+16=0 ?
The solutions to the equation -4x^2-20x+16=0 are x=-(5+sqrt(41))/2 ,x=(sqrt(41)-5)/2
Find the zeros of -4x^2-20x+16=0
The zeros of -4x^2-20x+16=0 are x=-(5+sqrt(41))/2 ,x=(sqrt(41)-5)/2