What are the solutions to the equation -16x^2+40x+62=0 ?

The solutions to the equation -16x^2+40x+62=0 are x=-(-5+sqrt(87))/4 ,x=(5+sqrt(87))/4

Find the zeros of -16x^2+40x+62=0

The zeros of -16x^2+40x+62=0 are x=-(-5+sqrt(87))/4 ,x=(5+sqrt(87))/4

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- 16 x^{2} + 40 x + 62 = 0

x = - \frac{- 5 + 87}{4}, x = \frac{5 + 87}{4}

Decimal Notation

x = - 1.08184 \dots, x = 3.58184 \dots

Solution steps

- 16 x^{2} + 40 x + 62 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 ( - 16 ) \cdot 62}{2 ( - 16 )}

4 0^{2} - 4 (- 16) \cdot 62 = 887

x_{1, 2} = \frac{- 40 \pm 8 87}{2 ( - 16 )}

Separate the solutions

x_{1} = \frac{- 40 + 8 87}{2 ( - 16 )}, x_{2} = \frac{- 40 - 8 87}{2 ( - 16 )}

x = \frac{- 40 + 8 87}{2 ( - 16 )} : - \frac{- 5 + 87}{4}

x = \frac{- 40 - 8 87}{2 ( - 16 )} : \frac{5 + 87}{4}

The solutions to the quadratic equation are:

x = - \frac{- 5 + 87}{4}, x = \frac{5 + 87}{4}

4 x^{2} + (n - 6) x - 36 = 0

u^{2} - 6 u - 7 = 0

x^{2} - x - 2 = 28

y^{2} - 14 y + 49 = 9

\frac{m ^{2}}{4} - 2 m + 4 = 0

What are the solutions to the equation -16x^2+40x+62=0 ?
The solutions to the equation -16x^2+40x+62=0 are x=-(-5+sqrt(87))/4 ,x=(5+sqrt(87))/4
Find the zeros of -16x^2+40x+62=0
The zeros of -16x^2+40x+62=0 are x=-(-5+sqrt(87))/4 ,x=(5+sqrt(87))/4