What are the solutions to the equation -4x^2-42x+22=0 ?

The solutions to the equation -4x^2-42x+22=0 are x=-11,x= 1/2

Find the zeros of -4x^2-42x+22=0

The zeros of -4x^2-42x+22=0 are x=-11,x= 1/2

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- 4 x^{2} - 42 x + 22 = 0

x = - 11, x = \frac{1}{2}

Decimal Notation

x = - 11, x = 0.5

Solution steps

- 4 x^{2} - 42 x + 22 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 42 ) \pm ( - 42 ) ^{2} - 4 ( - 4 ) \cdot 22}{2 ( - 4 )}

(- 42)^{2} - 4 (- 4) \cdot 22 = 46

x_{1, 2} = \frac{- ( - 42 ) \pm 46}{2 ( - 4 )}

Separate the solutions

x_{1} = \frac{- ( - 42 ) + 46}{2 ( - 4 )}, x_{2} = \frac{- ( - 42 ) - 46}{2 ( - 4 )}

x = \frac{- ( - 42 ) + 46}{2 ( - 4 )} : - 11

x = \frac{- ( - 42 ) - 46}{2 ( - 4 )} : \frac{1}{2}

The solutions to the quadratic equation are:

x = - 11, x = \frac{1}{2}

v^{2} + 3 v - 28 = 0

x^{2} - 13 x + 36 = 0

30 v = 9 v^{2} + 25

x^{2} - 2 (m^{2} - 4 m) x + m^{4} = 0

x, x^{2} + 5 x y = 9 x + 3 y

What are the solutions to the equation -4x^2-42x+22=0 ?
The solutions to the equation -4x^2-42x+22=0 are x=-11,x= 1/2
Find the zeros of -4x^2-42x+22=0
The zeros of -4x^2-42x+22=0 are x=-11,x= 1/2