What are the solutions to the equation -8x^2-62x+16=0 ?

The solutions to the equation -8x^2-62x+16=0 are x=-8,x= 1/4

Find the zeros of -8x^2-62x+16=0

The zeros of -8x^2-62x+16=0 are x=-8,x= 1/4

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- 8 x^{2} - 62 x + 16 = 0

x = - 8, x = \frac{1}{4}

Decimal Notation

x = - 8, x = 0.25

Solution steps

- 8 x^{2} - 62 x + 16 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 62 ) \pm ( - 62 ) ^{2} - 4 ( - 8 ) \cdot 16}{2 ( - 8 )}

(- 62)^{2} - 4 (- 8) \cdot 16 = 66

x_{1, 2} = \frac{- ( - 62 ) \pm 66}{2 ( - 8 )}

Separate the solutions

x_{1} = \frac{- ( - 62 ) + 66}{2 ( - 8 )}, x_{2} = \frac{- ( - 62 ) - 66}{2 ( - 8 )}

x = \frac{- ( - 62 ) + 66}{2 ( - 8 )} : - 8

x = \frac{- ( - 62 ) - 66}{2 ( - 8 )} : \frac{1}{4}

The solutions to the quadratic equation are:

x = - 8, x = \frac{1}{4}

(4 + 3 y)^{2} + y^{2} = 146

x, x^{2} y - x = 4 y

- x^{2} + 4 x = 0

4 x^{2} + 12 x + 3 = 0

2 (2 x - 3)^{2} - 2 (2 x - 3) - 2 = 0

What are the solutions to the equation -8x^2-62x+16=0 ?
The solutions to the equation -8x^2-62x+16=0 are x=-8,x= 1/4
Find the zeros of -8x^2-62x+16=0
The zeros of -8x^2-62x+16=0 are x=-8,x= 1/4