What are the solutions to the equation -4x^2-7x+12=0 ?

The solutions to the equation -4x^2-7x+12=0 are x=-(7+sqrt(241))/8 ,x=(sqrt(241)-7)/8

Find the zeros of -4x^2-7x+12=0

The zeros of -4x^2-7x+12=0 are x=-(7+sqrt(241))/8 ,x=(sqrt(241)-7)/8

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- 4 x^{2} - 7 x + 12 = 0

x = - \frac{7 + 241}{8}, x = \frac{241 - 7}{8}

Solution steps

- 4 x^{2} - 7 x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 ( - 4 ) \cdot 12}{2 ( - 4 )}

(- 7)^{2} - 4 (- 4) \cdot 12 = 241

x_{1, 2} = \frac{- ( - 7 ) \pm 241}{2 ( - 4 )}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 241}{2 ( - 4 )}, x_{2} = \frac{- ( - 7 ) - 241}{2 ( - 4 )}

x = \frac{- ( - 7 ) + 241}{2 ( - 4 )} : - \frac{7 + 241}{8}

x = \frac{- ( - 7 ) - 241}{2 ( - 4 )} : \frac{241 - 7}{8}

The solutions to the quadratic equation are:

x = - \frac{7 + 241}{8}, x = \frac{241 - 7}{8}

s^{2} - 4 s + 8 = 0

x^{2} + 10 x = 56

x^{2} - 18 x + 9 = 0

x^{2} + 49 - 14 x + x^{2} = 25

c, x^{2} - x y + y^{2} = c^{2}

What are the solutions to the equation -4x^2-7x+12=0 ?
The solutions to the equation -4x^2-7x+12=0 are x=-(7+sqrt(241))/8 ,x=(sqrt(241)-7)/8
Find the zeros of -4x^2-7x+12=0
The zeros of -4x^2-7x+12=0 are x=-(7+sqrt(241))/8 ,x=(sqrt(241)-7)/8